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Question: If \(\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\) and \(\cos \left( A-B \right)=\cos A\cos B+...

If sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B and cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B , find the values of sin⁡15\sin 15 and cos⁡15\cos 15 .

Explanation

Solution

We will use the above two formula sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B orcos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B, then we will put the values of A = 45 and B = 30, then we will substitute the values in the given formula and find the value of sin15 and cos15.

Complete step-by-step answer:
Let’s start our solution,
Putting the values of A = 45 and B = 30 in sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B we get,
sin⁡(45−30)=sin⁡45cos⁡30−cos⁡45sin⁡30\sin \left( 45-30 \right)=\sin 45\cos 30-\cos 45\sin 30
Now we know that, sin⁡30=12\sin 30=\dfrac{1}{2} , sin⁡45=cos⁡45=12\sin 45=\cos 45=\dfrac{1}{\sqrt{2}} and cos⁡30=32\cos 30=\dfrac{\sqrt{3}}{2} using this we get,
sin⁡15=12×32−12×12 sin⁡15=3−122 \begin{aligned} & \sin 15=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2} \\\ & \sin 15=\dfrac{\sqrt{3}-1}{2\sqrt{2}} \\\ \end{aligned}
Hence, we have found the value of sin15 using the formula sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B
Now we will use the formula cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B to find the value of cos15
Putting the values of A = 45 and B = 30 in cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B we get,
cos⁡(45−30)=cos⁡45cos⁡30+sin⁡45sin⁡30\cos \left( 45-30 \right)=\cos 45\cos 30+\sin 45\sin 30
Now we know that, sin⁡30=12\sin 30=\dfrac{1}{2} , sin⁡45=cos⁡45=12\sin 45=\cos 45=\dfrac{1}{\sqrt{2}} and cos⁡30=32\cos 30=\dfrac{\sqrt{3}}{2} using this we get,
cos⁡15=12×32+12×12 cos⁡15=3+122 \begin{aligned} & \cos 15=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2} \\\ & \cos 15=\dfrac{\sqrt{3}+1}{2\sqrt{2}} \\\ \end{aligned}
Hence, we have found the value of cos15 using the formula cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B
Hence we have found the value of both sin15 and cos15.

Note: Here we have used the given trigonometric formula to find the value of sin15 and cos15. One can also use the formula cos⁡2x=2cos⁡2x−1\cos 2x=2{{\cos }^{2}}x-1 and the put the value of x = 15 and find the value of cos15 from there and then use the relation sin⁡x=1−cos⁡2x\sin x=\sqrt{1-{{\cos }^{2}}x}then substitute the value of cos15 and find the value of sin15 from there. In this method we have to do less calculation compared to the one that is given in the solution, but the formulas to use must be remembered.