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Home » If sin A = 3/5, cos B = –12/13, where A and B, both lie in second quadrant, find the value of sin (A +B).
If sin A = 3/5, cos B = –12/13, where A and B, both lie in second quadrant, find the value of sin (A +B).
CBSE | Class 11 | Exercise 7.1 | Maths | RD Sharma | Values of Trigonometric Functions at Sum or Difference of Angles
If sin A = 3/5, cos B = –12/13, where A and B, both lie in second quadrant, find the value of sin (A +B).

As per the question given:

    \[sin\text{ }A\text{ }=\text{ }3/5,\text{ }cos\text{ }B\text{ }=-\text{ }12/13,\]

    \[A\text{ }and\text{ }B,\]

both lie in second quadrant.

Since,

    \[cos\text{}A\text{}=\text{}\text{}\surd(1\text{}-\text{}si{{n}^{2}}~A)\text{}and\text{}sin\text{}B\text{}=\text{}\surd (1\text{}-\text{}co{{s}^{2}}~B)\]

Let’s get the value of

    \[cos\text{ }A\text{ }and\text{ }sin\text{ }B\]

    \[cos\text{ }A\text{ }=\text{ }\text{ }\surd (1\text{ }-\text{ }si{{n}^{2}}~A)\]

    \[=\text{ }\text{ }\surd (1\text{ }-\text{ }{{\left( 3/5 \right)}^{2}})\]

    \[=\text{ }\text{ }\surd \left( 1-\text{ }9/25 \right)\]

    \[=\text{ }\text{ }\surd \left( \left( 25-9 \right)/25 \right)\]

    \[=\text{ }\text{ }\surd \left( 16/25 \right)\]

    \[=\text{ }\text{ }4/5\]

And,

    \[sin\text{ }B\text{ }=\text{ }\surd (1\text{ }-\text{ }co{{s}^{2}}~B)\]

    \[=\text{ }\surd (1\text{ }-\text{ }{{\left( -12/13 \right)}^{2}})\]

    \[=\text{ }\surd \left( 1\text{ }-\text{ }144/169 \right)\]

    \[=\text{ }\surd \left( \left( 169-144 \right)/169 \right)\]

    \[=\text{ }\surd \left( 25/169 \right)\]

    \[=\text{ }5/13\]

 

We have to find

    \[sin\text{ }\left( A\text{ }+\text{ }B \right)\]

Since,

    \[sin\text{ }\left( A\text{ }+\text{ }B \right)\text{ }=\text{ }sin\text{ }A\text{ }cos\text{ }B\text{ }+\text{ }cos\text{ }A\text{ }sin\text{ }B\]

    \[=\text{ }3/5\text{ }\times \text{ }\left( -12/13 \right)\text{ }+\text{ }\left( -4/5 \right)\text{ }\times \text{ }5/13\]

    \[=\text{ }-36/65\text{ }-\text{ }20/65\]

    \[=\text{ }-56/65\]

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