Solveeit Logo
Login

Question

Question: If \(A+B+C={{270}^{\circ }}\), then \(\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C\) is equal to (a...

If A+B+C=270∘A+B+C={{270}^{\circ }}, then cos⁡2A+cos⁡2B+cos⁡2C+4sin⁡Asin⁡Bsin⁡C\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C is equal to
(a)0
(b)1
(c)2
(d)3

Explanation

Solution

Use the following trigonometric identities to simplify the given equation.
cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A+B)=\cos A\cos B-\sin A\sin B
cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos (A-B)=\cos A\cos B+\sin A\sin B
cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)
cos⁡A−cos⁡B=2sin⁡(A+B2)sin⁡(A−B2)\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)
cos⁡2A=1−2sin⁡2A\cos 2A=1-2{{\sin }^{2}}A

Complete step by step answer:
We have, if A+B+C=270∘A+B+C={{270}^{\circ }}, then cos⁡2A+cos⁡2B+cos⁡2C+4sin⁡Asin⁡Bsin⁡C=?\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C=?
Let cos⁡2A+cos⁡2B+cos⁡2C+4sin⁡Asin⁡Bsin⁡C=k\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C=k.
Now,
k=cos⁡2A+cos⁡2B+cos⁡2C+4sin⁡Asin⁡Bsin⁡C...................(1)k=\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C...................(1)
We know the trigonometric identity, cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)
Using this trigonometric identity, equation (1) becomes,
k=cos⁡2A+(cos⁡2B+cos⁡2C)+4sin⁡Asin⁡Bsin⁡Ck=\cos 2A+(\cos 2B+\cos 2C)+4\sin A\sin B\sin C
=cos⁡2A+(2cos⁡(2B+2C2)cos⁡(2B−2C2))+4sin⁡Asin⁡Bsin⁡C=\cos 2A+\left( 2\cos \left( \dfrac{2B+2C}{2} \right)\cos \left( \dfrac{2B-2C}{2} \right) \right)+4\sin A\sin B\sin C
=cos⁡2A+(2cos⁡(B+C)cos⁡(B−C))+4sin⁡Asin⁡Bsin⁡C=\cos 2A+\left( 2\cos \left( B+C \right)\cos \left( B-C \right) \right)+4\sin A\sin B\sin C
We are given that, A+B+C=270∘A+B+C={{270}^{\circ }}.
On rearranging, the above can be rewritten as B+C=270∘−AB+C={{270}^{\circ }}-A.
Substituting B+C=270∘−AB+C={{270}^{\circ }}-A in the equation, we obtain,
k=cos⁡2A+2cos⁡(270∘−A)cos⁡(B−C)+4sin⁡Asin⁡Bsin⁡Ck=\cos 2A+2\cos \left( {{270}^{\circ }}-A \right)\cos \left( B-C \right)+4\sin A\sin B\sin C
We know that, cos⁡(270∘−θ)=−sin⁡θ\cos ({{270}^{\circ }}-\theta )=-\sin \theta (a proof for the same is discussed later on in this article)
Hence,
k=cos⁡2A−2sin⁡Acos⁡(B−C)+4sin⁡Asin⁡Bsin⁡Ck=\cos 2A-2\sin A\cos \left( B-C \right)+4\sin A\sin B\sin C
We also know that, cos⁡2A=1−2sin⁡2A\cos 2A=1-2{{\sin }^{2}}A
Substituting cos⁡2A=1−2sin⁡2A\cos 2A=1-2{{\sin }^{2}}A in the equation, we obtain,
k=1−2sin⁡2A−2sin⁡Acos⁡(B−C)+4sin⁡Asin⁡Bsin⁡Ck=1-2{{\sin }^{2}}A-2\sin A\cos \left( B-C \right)+4\sin A\sin B\sin C
We take 2sin⁡A2\sin A common for the second and third term. The equation then becomes,
k=1−2sin⁡A(sin⁡A+cos⁡(B−C))+4sin⁡Asin⁡Bsin⁡Ck=1-2\sin A\left( \sin A+\cos \left( B-C \right) \right)+4\sin A\sin B\sin C
We are given that, A+B+C=270∘A+B+C={{270}^{\circ }}.
On rearranging, the above can be rewritten as A=270∘−(B+C)A={{270}^{\circ }}-(B+C).
Substituting A=270∘−(B+C)A={{270}^{\circ }}-(B+C) in the equation, we get,
k=1−2sin⁡A(sin⁡(270∘−(B+C))+cos⁡(B−C))+4sin⁡Asin⁡Bsin⁡Ck=1-2\sin A\left( \sin \left( {{270}^{\circ }}-(B+C) \right)+\cos \left( B-C \right) \right)+4\sin A\sin B\sin C
We know that, sin⁡(270∘−θ)=−cos⁡θ\sin ({{270}^{\circ }}-\theta )=-\cos \theta
Hence, the equation can be rewritten as,
k=1−2sin⁡A(−cos⁡(B+C)+cos⁡(B−C))+4sin⁡Asin⁡Bsin⁡Ck=1-2\sin A\left( -\cos \left( B+C \right)+\cos \left( B-C \right) \right)+4\sin A\sin B\sin C
k=1−2sin⁡A(cos⁡(B−C)−cos⁡(B+C))+4sin⁡Asin⁡Bsin⁡Ck=1-2\sin A\left( \cos \left( B-C \right)-\cos \left( B+C \right) \right)+4\sin A\sin B\sin C
We know the trigonometric identity, cos⁡A−cos⁡B=2sin⁡(A+B2)sin⁡(A−B2)\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)
Using the above trigonometric identity, the equation can be rewritten as,
k=1−2sin⁡A(2sin⁡(B+C+B−C2)sin⁡(B+C−B+C2))+4sin⁡Asin⁡Bsin⁡Ck=1-2\sin A\left( 2\sin \left( \dfrac{B+C+B-C}{2} \right)\sin \left( \dfrac{B+C-B+C}{2} \right) \right)+4\sin A\sin B\sin C
k=1−2sin⁡A(2sin⁡(2B2)sin⁡(2C2))+4sin⁡Asin⁡Bsin⁡Ck=1-2\sin A\left( 2\sin \left( \dfrac{2B}{2} \right)\sin \left( \dfrac{2C}{2} \right) \right)+4\sin A\sin B\sin C
k=1−2sin⁡A.2sin⁡Bsin⁡C+4sin⁡Asin⁡Bsin⁡Ck=1-2\sin A.2\sin B\sin C+4\sin A\sin B\sin C
k=1−4sin⁡Asin⁡Bsin⁡C+4sin⁡Asin⁡Bsin⁡Ck=1-4\sin A\sin B\sin C+4\sin A\sin B\sin C
∴k=1\therefore k=1
But, from equation (1),
k=cos⁡2A+cos⁡2B+cos⁡2C+4sin⁡Asin⁡Bsin⁡Ck=\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C
∴cos⁡2A+cos⁡2B+cos⁡2C+4sin⁡Asin⁡Bsin⁡C=1\therefore \cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C=1
Hence, the value of cos⁡2A+cos⁡2B+cos⁡2C+4sin⁡Asin⁡Bsin⁡C\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin Cat A+B+C=270∘A+B+C={{270}^{\circ }} is equal to 1.

So, the correct answer is “Option B”.

Note: Remembering a lot of trigonometric equations may seem tricky and confusing. But, we can easily derive any of the trigonometric equations from the very basic trigonometric equations.
For example,
cos⁡(270∘−θ)=−sin⁡θ\cos \left( {{270}^{\circ }}-\theta \right)=-\sin \theta . One can easily go wrong with signs of such equations. If we look at the LHS of this equation we can see that it is in the form of the trigonometric identitycos⁡(A−B)\cos (A-B). Using the this trigonometric equation cos⁡(270∘−θ)\cos ({{270}^{\circ }}-\theta ) can be written as cos⁡270∘cos⁡θ+sin⁡270∘sin⁡θ\cos {{270}^{\circ }}\cos \theta +\sin {{270}^{\circ }}\sin \theta .
Value of cos⁡270∘=0\cos {{270}^{\circ }}=0 and value of sin⁡270∘=−1\sin {{270}^{\circ }}=-1
Hence, cos⁡(270∘−θ)=(−1)sin⁡θ=−sin⁡θ\cos ({{270}^{\circ }}-\theta )=(-1)\sin \theta =-\sin \theta