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Question: If \(\sin x=\dfrac{3}{5}\) , \(\cos y=-\dfrac{12}{13}\) , where x and y lies in the second quadrant,...

If sin⁡x=35\sin x=\dfrac{3}{5} , cos⁡y=−1213\cos y=-\dfrac{12}{13} , where x and y lies in the second quadrant, find the value of sin⁡(x+y)\sin \left( x+y \right) .

Explanation

Solution

Hint: Start by finding the value of cosx and siny using the relation that the sum of squares of cosine and sine function is equal to 1. Now once you have got the value of cosx and siny, you can easily find the required answer using the formula sin(x+y)=sinxcosy+cosxsiny.

Complete step-by-step answer:

We know that cos⁡2x=1−sin⁡2x.{{\cos }^{2}}x=1-{{\sin }^{2}}x. So, if we put the value of sinx in the formula, we get
cos⁡2x=1−(35)2{{\cos }^{2}}x=1-{{\left( \dfrac{3}{5} \right)}^{2}}
⇒cos⁡2x=1−925\Rightarrow {{\cos }^{2}}x=1-\dfrac{9}{25}
⇒cos⁡2x=1625\Rightarrow {{\cos }^{2}}x=\dfrac{16}{25}
Now we know that a2=b{{a}^{2}}=b implies a=±ba=\pm \sqrt{b} . So, our equation becomes:
⇒cos⁡x=±1625=±45\Rightarrow \cos x=\pm \sqrt{\dfrac{16}{25}}=\pm \dfrac{4}{5}
Now, as it is given that x lies in the second quadrant and cosx is negative in the second quadrant.
∴cos⁡x=−45\therefore \cos x=-\dfrac{4}{5}
We also know that sin⁡2y=1−cos⁡2y.{{\sin }^{2}}y=1-{{\cos }^{2}}y. So, if we put the value of cosy in the formula, we get
sin⁡2y=1−(−1213)2{{\sin }^{2}}y=1-{{\left( -\dfrac{12}{13} \right)}^{2}}
⇒sin⁡2y=1−144169\Rightarrow {{\sin }^{2}}y=1-\dfrac{144}{169}
⇒sin⁡2y=25169\Rightarrow {{\sin }^{2}}y=\dfrac{25}{169}
Now we know that a2=b{{a}^{2}}=b implies a=±ba=\pm \sqrt{b} . So, our equation becomes:
⇒siny⁡=±25169=±513\Rightarrow \operatorname{siny}=\pm \sqrt{\dfrac{25}{169}}=\pm \dfrac{5}{13}
Now, as it is given that y lies in the second quadrant and siny is positive in the second quadrant.
∴siny⁡=513\therefore \operatorname{siny}=\dfrac{5}{13}
Now we know that sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y . If we put the required values in the formula, we get
sin⁡(x+y)=35×(−1213)+(−45)×513=−36−2065=−5665\sin \left( x+y \right)=\dfrac{3}{5}\times \left( -\dfrac{12}{13} \right)+\left( -\dfrac{4}{5} \right)\times \dfrac{5}{13}=\dfrac{-36-20}{65}=\dfrac{-56}{65}
So, the value of sin(x+y) is −5665-\dfrac{56}{65} .

Note: It is useful to remember the graph of the trigonometric ratios along with the signs of their values in different quadrants. For example: sine is always positive in the first and the second quadrant while negative in the other two. Also, you need to remember the properties related to complementary angles and trigonometric ratios.