Answer:
Prove that
Answer: = cosx cos2x cos4x cos8x Multiply and divide by 2sinx, we get We know that, sin 2x = 2 sinx cosx Replacing x by 2x, we get sin 2(2x) = 2 sin(2x) cos(2x) or sin 4x = 2 sin 2x cos 2x...
Prove that
Answer: Taking sinx common from the numerator and cosx from the denominator
Prove that:
Answer: = RHS ∴ LHS = RHS Hence Proved
Prove that:
Answer: Multiply and divide by 2, we get = RHS ∴ LHS = RHS Hence Proved
Prove that cot x – 2cot 2x = tan x
Answer: Taking LHS, = cot x – 2cot 2x …(i) We know that, cot x = cos x/ sin x Replacing x by 2x, we get cot 2x = cos 2x/ sin 2x So, eq. (i) becomes = sin x/ cos x = tan x = RHS ∴ LHS = RHS Hence...
Prove that cos 2x + 2sin2 x = 1
Answer: Taking LHS = 2 – 1 = 1 = RHS
Prove that cosec 2x + cot 2x = cot x
Answer: To Prove: cosec 2x + cot 2x = cot x Taking LHS, = cosec 2x + cot 2x …(i) We know that, cosec x = 1 / sin x and cot x = cos x/ sin x Replacing x by 2x, we get = cos x/ sinx = cot...
Prove that sin 2x(tan x + cot x) = 2
Answer: Taking LHS, sin 2x(tan x + cot x) We know that: We know that, sin 2x = 2 sinx cosx = 2 = RHS ∴ LHS = RHS Hence Proved
Prove that:
Answer: = sin x / cos x = tan x = RHS ∴ LHS = RHS Hence Proved
Prove that:
Answer: = cot x = RHS ∴ LHS = RHS Hence Proved
Prove that:
Answer: = tan x = RHS ∴ LHS = RHS Hence Proved
Prove that:
Answer: Taking LHS = cos x + sin x = RHS ∴ LHS = RHS Hence Proved
If cos x = -1/3 , find the value of cos 3x
Answer; We know that, cos 3x = 4cos3 x – 3 cosx Putting the values, we get
If sinx = 1/6, find the value of sin 3x.
Answer: To find: sin 3x We know that, sin 3x = 3 sinx – sin3 x Putting the values, we get
If , find the values of tan 2x
Answer: We know that:
If , find the values of cos 2x
Answer:
If , find the values of sin 2x
Answer: We know that
If , find the values of tan 2x
Answer: Replacing x by 2x, we get tan 2x = sin 2x / cos 2x Putting the values of sin 2x and cos 2x, we get
If , find the values of cos 2x
Answer: We know that, cos 2x = 2cos2 x – 1 Putting the value, we get
If , find the values of sin 2x
Answer: We know that, sin2x = 2 sinx cosx …(i) Here, we don’t have the value of sin x. So, firstly we have to find the value of sinx We know that, cos2 x + sin2 x = 1 Putting the values, we get...
If , find the values of tan 2x
Answer: We know that: Replacing x by 2x, we get tan 2x = sin 2x / cos 2x Putting the values of sin 2x and cos 2x, we get
If , find the values of cos 2x
Answer: We know that, cos 2x = 1 – 2sin2 x Putting the value, we get
If , find the values of sin 2x
Answer: Given: $\sin x=\frac{\sqrt{5}}{3}$ To find: sin2x We know that, sin2x = 2 sinx cosx …(i) Here, we don’t have the value of cos x. So, firstly we have to find the value of cosx We know that,...
Prove that
Answer:
Prove that:
Answer: Taking LHS
Prove that:
Answer:
Prove that:
Answer: Taking LHS
If , prove that
Answer: cosx + cosy = 1/3--------- i sinx + siny = 1/3----------ii dividing ii by i we get Using the formula, sinA + sinB = 2sin(A+B)/2 . cos(A-B)/2 cosA + cosB = 2cos(A+B)/2 ....
Prove that:
Answer; Taking LHS = 3/16
Prove that:
= 3/16
Prove that:
Answer: = 1/16
Prove that:
Answer: = cot 5x Using the formulas, 2cosAsinB = sin (A + B) – sin (A - B) 2sinAsinB = cos (A - B) – cos (A + B)
Prove that:
Answer: = tan 2x Using the formulas, 2cosAsinB = sin (A + B) – sin (A - B) 2cosAcosB = cos (A + B) + cos (A - B) 2sinAsinB = cos (A - B) – cos (A + B)
Prove that:
Answer: =Sin3x+sin2x-sinx = (sin3x- sinx)+sin2x = ( 2cos(3x+x)/2 . sin(3x-x)/2 ) + sin2x = 2cos2xsinx +sin2x = 2cos2xsinx + 2sinxcosx = 2sinx (cos2x + cosx ) = 2sinx (2cos(2x+x)/2 . cos(2x-x)/2 ) =...
Prove that:
Answer: Using the formula, sinA + sinB = 2sin(A+B)/2 . cos(A-B)/2 sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2
Prove that:
Answer: Using the formula, cosA - cosB = -2sin(A+B)/2 . sin(A-B)/2 cosA + cosB = 2cos(A+B)/2 . cos(A-B)/2
Prove that:
Answer: Using the formula, cosA - cosB = -2sin(A+B)/2 . sin(A-B)/2 sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2
Prove that:
Using the formula, cosA - cosB = -2sin(A+B)/2 . sin(A-B)/2 sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2
Prove that (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
Answer: = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = (2sin(3x+x)/2 . cos(3x-x)/2) sin x + (-2sin(3x+x)/2 . sin(3x-x)/2) cosx = (2sin2x cosx) sinx-(2sin2x sinx) cosx = 0. Using the formula,...
Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Answer: L.H.S cot 4x (sin 5x + sin3x) = cot 4x (2sin(5x+3x)/2 . cos (5x-3x)/2 ) = cot 4x (2 sin4x cosx) = (cos4x / sin 4x).(2 sin4x cosx) = 2cos4xcosx R.H.S cot x (sin 5x - sin3x) = cot x...
Prove that:
Answer: = sin 6x / cos 6x = tan 6x Using the formula, sinA + sinB = 2sin(A+B)/2 . cos(A-B)/2 cosA + cosB = 2cos(A+B)/2 . cos(A-B)/2
Prove that:
Answer: Using the formula, cosA - cosB = -2sin(A+B)/2 . sin(A-B)/2 sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2
Prove that:
Answer: = tan4x Using the formula, sinA + sinB = 2sin(A+B)/2 . cos(A-B)/2 cosA + cosB = 2cos(A+B)/2 . cos(A-B)/2
Prove that:
Answer: = sin x / cos x = tanx Using the formula, sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2 cosA + cosB = 2cos(A+B)/2 . cos(A-B)/2
Prove that
Answer: = cos x / sin x = cot x Using the formula, sinA + sinB = 2sin (A+B)/2 . cos (A-B)/2 cosA - cosB = -2sin (A+B)/2 . sin (A-B)/2
Express each of the following as an algebraic sum of sines or cosines : (iii) 2cos 7x cos 3x (iv) 2sin 8x sin 2x
Answer: iii) 2cos7xcos3x = cos (7x+3x) + cos (7x – 3x) =cos10x + cos 4x Using, 2cosAcosB = cos (A+ B) + cos (A - B) iv)2sin8xsin2x = cos (8x - 2x) – cos (8x + 2x) = cos6x – cos10x Using, 2sinAsinB =...
Express each of the following as an algebraic sum of sines or cosines : (i) 2sin 6x cos 4x (ii) 2cos 5x din 3x
Answer: i) 2sin 6x cos 4x = sin (6x+4x) + sin (6x-4x) = sin 10x + sin 2x Using, 2sinAcosB = sin (A+ B) + sin (A - B) ii) 2cos 5x sin3x = sin (5x + 3x) – sin (5x – 3x) = sin8x – sin2x Using,...
Express each of the following as a product. 3. cos 7x + cos 5x 4. cos2x – cos 4x
Answer:
Express each of the following as a product. 1. sin 10x + sin 6x 2. sin 7x – sin 3x
Answer:
Prove that:
Prove that
Answer: In this question the following formulas will be used:
Prove that
Answer: In this question the following formulas will be used: sin (A - B) = sinA cos B - cosA sinB cos (A - B) = cosAcosB+ sinAsinB
Prove that cos x + cos (120 – x) + cos (120 + x) = 0
Answer: In this question the following formulas will be used: cos (A + B) = cosAcosB – sinAsinB cos (A - B) = cosAcosB+ sinAsinB = cos x + cos 120 cosx – sin120sinx + cos 120 cosx+sin120sinx = cosx...
Prove that sin(150 + x) + sin (150 – x) = cos x
Answer: In this question the following formula will be used: Sin( A +B)= sinA cos B + cosA sinB Sin( A - B)= sinA cos B - cosA sinB = sin150 cosx + cos 150 sinx + sin150 cosx – cos150 sinx =2sin150...
Prove that:
Answer:
Prove that:
Answer:
Prove that:
Answer:
Prove that:
Answer:
If , where , find the values of (iii) tan (x + y)
Answer: (iii)Here first we will calculate value of tanx and tany,
If , where , find the values of (i) sin (x + y) (ii) cos (x – y)
Answer: Given: $\cos x=\frac{3}{5}\,and\,\cos y=-\frac{24}{25}$ We will first find out value of sinx and siny, (i)sin(x + y) = sinx.cosy + cosx.siny (ii)cos(x - y) = cosx.cosy +...
If , where , find the values of (iii) tan (x – y)
Answer: Given: $\sin x=\frac{12}{13}\,and\,\sin y=\frac{4}{5}$ (iii)Here first we will calculate value of tanx and tany
If , where , find the values of (i) sin (x + y) (ii) cos (x + y)
Answer: Given: $\sin x=\frac{12}{13}\,and\,\sin y=\frac{4}{5}$ Here we will find values of cosx and cosy (i) sin(x + y) = sinx.cosy + cosx.siny (ii)cos(x + y) = cosx.cosy +...
If x and y are acute angles such that , prove that
Answer: Given: $\cos x=\frac{13}{14}\,and\,\cos y=\frac{1}{7}$ Now we will calculate value of sinx and siny Hence, Cos(x - y) = cosx.cosy + sinx.siny
If x and y are acute such that , prove that
Answer: Given: $\sin x=\frac{1}{\sqrt{5}}\,and\,\sin y=\frac{1}{\sqrt{10}}$ Now we will calculate value of cos x and cosy Sin(x + y) = sinx.cosy + cosx.siny
If θ and Φ lie in the first quadrant such that , find the values of (iii) tan (θ – Φ)
Answer: (iii)We will first find out the Values of tanθ and tanΦ
If θ and Φ lie in the first quadrant such that , find the values of (i) sin (θ – Φ ) (ii) cos (θ – Φ)
Answer: Given: $\sin \theta =\frac{8}{17}\,and\,\cos \phi =\frac{12}{13}$
Prove that:
Answer: Using cos(90° + θ) = - sinθ(I quadrant cosx is positive cosec( - θ) = - cosecθ tan(270° - θ) = tan(180° + 90° - θ) = tan(90° - θ) = cotθ (III quadrant tanx is positive) Similarily sin(270° +...
Prove that:
Answer: Using sin(90° + θ) = cosθ and sin( - θ) = sinθ,tan(90° + θ) = - cotθ Sin(180° + θ) = - sinθ(III quadrant sinx is negative)
Prove that
Answer:
Prove that:
Answer:
Prove that:
Answer:
Prove that:
Answer: (II quadrant tanx negative) - tan45° = - 1
Prove that
Answer: (ii)cot105° - tan105° = cot(180° - 75°) - tan(180° - 75°) (II quadrant tanx is negative and cotx as well) = - cot75° - ( - tan75°) = tan75° - cot75°
Prove that: tan15° + cot15° = 4
Answer: (iii) tan15° + cot15° = First, we will calculate tan15°,
Prove that:
Answer: (i) sin75° = sin(90° - 15°) .…….(using sin(A - B) = sinAcosB - cosAsinB) = sin90°cos15° - cos90°sin15° = 1.cos15° - 0.sin15° = cos15° Cos15° = cos(45° - 30°) …………(using cos(A - B) = cosAcosB...
Prove that:
Answer:
Prove that:
(i) cos(n + 2)x.cos(n + 1)x + sin(n + 2)x.sin(n + 1)x = cos x Answer: (i) cos(n + 2)x.cos(n + 1)x + sin(n + 2)x.sin(n + 1)x = sin((n + 2)x + (n + 1)x)(using cos(A - B) = cosAcosB + sinAsinB) =...
Prove that (i) sin(50° + θ)cos(20° + θ) – cos(50° + θ)sin(20° + θ) = 1/2 (ii) cos(70° + θ)cos(10° + θ) + sin(70° + θ)sin(10° + θ) = 1/2
Answer: (i) We have: sin(50° + θ)cos(20° + θ) - cos(50° + θ)sin(20° + θ) = sin(50° + θ - (20° + θ))(using sin(A - B) = sinAcosB - cosAsinB) = sin(50° + θ - 20° - θ) = sin30° = 1/2 (ii) We have:...
Prove that (v) cos130°cos40° + sin130°sin40° = 0
(v) cos130°cos40° + sin130°sin40° = cos(130° - 40°) (using cos(A - B) = cosAcosB + sinAsinB) = cos90° = 0
Prove that
Answer: (iii) cos75°cos15° + sin75°sin15° = cos(75° - 15°) (using cos(A - B) = cosAcosB + sinAsinB) = cos60° = 1/2 (iv) sin40°cos20° + cos40°sin20° = sin(40° + 20°) (using sin(A + B) = sinAcosB +...
Prove that
Answer: (i) sin80°cos20° - cos80°sin20° = sin(80° - 20°) (using sin(A - B) = sinAcosB - cosAsinB) = sin60° = $\frac{\sqrt{3}}{2}$ (ii)cos45°cos15° - sin45°sin15° = cos(45° + 15°) (using cos(A + B) =...
Find the values of all trigonometric functions of 135 deg
Answer:
Find the value of (ix) cos (495֯ )
(ix)cos495° = cos(360° + 135°) …………(using cos(360° + x) = cosx) = cos135° = cos(180° - 45°) ………….(using cos(180° - x) = - cosx) = - cos45° = - 1//2
Find the value of (vii) cot ( – 315֯ ) (viii) sin ( – 1230֯ )
Answer: vii) $ \cot \left( -{{315}^{\circ }} \right)=\frac{1}{\tan \left( -{{315}^{\circ }} \right)} $ $ \Rightarrow \frac{1}{-\tan \left( {{315}^{\circ }} \right)}=\frac{1}{-\tan \left(...
Find the value of (v) cosec ( – 690֯ ) (vi) tan (225֯ )
Answer:
Find the value of (iii) tan ( – 120֯ ) (iv) sec ( – 420֯ )
Answer: (iii) tan( - 120°) = - tan12 …….(tan( - x) = tanx) = - tan(180° - 60°) ……. (in II quadrant tanx is negative) = - ( - tan60°) = tan60°
Find the value of (i) cos 840֯(ii) sin 870֯
Answer: (i) Cos840° = Cos(2.360° + 120°) …………(using Cos(2ϖ + x) = Cosx) = Cos(120°) = Cos(180° - 60°) = - Cos60° ……………(using Cos(ϖ - x) = - Cosx) = - 1/2 (ii) sin870° = sin(2.360° + 150°)...
Prove that:
Answer: Taking LHS: We know that: Putting the values, we get = 4 = RHS ∴ LHS = RHS
Prove that
Answer: = RHS ∴ LHS = RHS
Prove that
Answer: Taking LHS, we have: = tan2 π/3 + 2 cos2 π/4 + 3 sec2 π/6 + 4 cos2 π/2 Putting π = 180° = tan2 60° + 2 cos2 45° + 3 sec2 30° + 4 cos2 90°Now, we know that, = 3 + 1 + 4 = 8 = RHS ∴ LHS =...
Find the value of cos (-2220 )
Answer: To find: Value of cos 2220° We have, cos (-2220 ) = cos 2220° [∵ cos(-θ) = cos θ] = cos [2160 + 60°] = cos [360° × 6 + 60°] = cos 60° [Clearly, 2220° is in I Quadrant and the multiple of...
Find the value of cosec (-750 )
Answer: To find: Value of cosec (-750°) We have, cosec (-750°) = - cosec(750°) [∵ cosec(-θ) = -cosec θ] = - cosec [90° × 8 + 30°] Clearly, 405° is in I Quadrant and the multiple of 90° is even = -...
Find the value of cot (585 )
Answer: We have, cot (585°) = cot [90° × 6 + 45°] = cot 45° [Clearly, 585° is in III Quadrant and the multiple of 90° is even] = 1 [∵ cot 45° = 1]
Find the value of tan (-300 )
Answer: To find: Value of tan (-300°) We have, tan (-300°) = - tan (300°) [∵ tan(-θ) = -tan θ] = - tan [90° × 3 + 30°] Clearly, 300° is in IV Quadrant and the multiple of 90° is odd = - cot 30° =...
Find the value of sec (-1470 )
Answer: To find: Value of sec (-1470°) We have, sec (-1470°) = sec (1470°) [∵ sec(-θ) = sec θ] = sec [90° × 16 + 30°] Clearly, 1470° is in I Quadrant and the multiple of 90° is even = sec 30° $\sec...
Find the value of sin 405°
Answer: To find: Value of sin 405° We have, sin 405° = sin [90° × 4 + 45°] = sin 45° [Clearly, 405° is in I Quadrant and the multiple of 90° is even] $\sin {{405}^{\circ...
Find the value of
Answer: We have: $ co\sec \left( -\frac{41\pi }{4} \right) $ We know that: $ co\sec \left( -\frac{41\pi }{4} \right)=-co\sec \left( \frac{41\pi }{4} \right) $ [∵ cosec(-θ) = -cosec θ] Putting π =...
Find the value of .
Answer: We have $\sec \left( -\frac{25\pi }{3} \right)$ We know that: $\sec \left( -\frac{25\pi }{3} \right)=\sec \left( \frac{25\pi }{3} \right)$ [∵ sec(-θ) = sec θ] Putting π = 180° = sec[25 ×...
Find the value of
Answer: We have: = cot (13 × 45°) = cot (585°) = cot [90° × 6 + 45°] = cot 45° [Clearly, 585° is in III Quadrant and the multiple of 90° is even] = 1 [∵ cot 45° = 1]
Find the value of .
Answer: We know that, tan(-θ) = - tan θ = - tan 60° = -√3 [∵ tan 60° = √3]
Find the value of
Answer:
Find the value of
Answer:
If , find the values of all the other five trigonometric functions.
Answer: Given: $\sec \,x=-2$ Given that: So, x lies in III Quadrant. So, sin and cos will be negative but tan will be positive. Now, we know that We know that, cos x + sin x = 1 Putting the values,...
If , find the value of sin x.
Answer: Given: $\cos \,x=-\frac{\sqrt{15}}{4}$ To find: value of sinx Given that: $\frac{\pi }{2}<x<\pi $ So, x lies in II quadrant, and sin will be positive. We know that, cos θ + sin θ = 1...
If and x lies in Quadrant III, find the values of cos x and cot x.
Answer: Given: $\sin \,x=-\frac{2\sqrt{6}}{5}$ Since, x is in III Quadrant. So, sin and cos will be negative but tan will be positive. We know that, sin x + cos x = 1 Putting the values, we get...
If and θ lies in Quadrant IV, find the values of all the other five trigonometric functions.
Given : $\sec \theta =\sqrt{2}$ Since, θ is in IV Quadrant. So, sin and tan will be negative but cos will be positive. Now, we know that We know that, cos θ + sin θ = 1 Putting the values, we get...
If and θ lies in Quadrant II, find the values of all the other five trigonometric functions.
Answer: Given: $\cos ec\,\theta =\frac{5}{3}$ Since, θ is in II Quadrant. So, cos and tan will be negative but sin will be positive. Now, we know that We know that, sin θ + cos θ = 1 Putting the...
If and θ lies in Quadrant IV, find the values of all the other five trigonometric functions.
Answer: Since, θ is in IV Quadrant. So, sin and tan will be negative but cos will be positive. We know that, sin θ + cos θ = 1 Putting the values, we get Since, θ in IV quadrant and cosθ is positive...
If and θ lies in Quadrant III, find the value of all the other five trigonometric functions.
Answer: We are given : $\cos \theta =-\frac{\sqrt{3}}{2}$ Since, θ is in III Quadrant. So, sin and cos will be negative but tan will be positive. We know that, cos θ + sin θ = 1 Putting the values,...
The function is the formula to convert x °C to Fahrenheit units. Find
(i) F(0),
(ii) F(–10),
(iii) The value of x when f(x) = 212. Interpret the result in each case.
Answer : Given: (i) To find: (i) F(0) Substituting the value of x = 0 in eq. (i), we get ⇒ F(0) = 32 It means 0° C = 32° F To find: (ii) F(-10) Substituting the value of x = -10 in eq. (i), we get...