Question

2sin^2 beta + 4cos(alpha +beta)sin alpha. sin beta + cos 2 ( alpha + beta)=​

Answer 1

Answer:

$Cos2\alpha$

Step-by-step explanation:

Here, we have to find out the value of

$2Sin^{2}\beta+4Cos(\alpha+\beta)Sin\alpha .Sin\beta +Cos2(\alpha+\beta)$

Now, we will calculate it.

$2Sin^{2}\beta+4Cos(\alpha+\beta)Sin\alpha .Sin\beta +Cos2(\alpha+\beta)$

=$(1-Cos2\beta)+4(Cos\alpha .Cos\beta -Sin\alpha .Sin\beta).Sin\alpha .Sin\beta+Cos2(\alpha+\beta)$

We have applied the following formulas:

$Cos2\beta=1-2Sin^{2}\beta$  ..... (1)and

$Cos(\alpha+\beta)=Cos\alpha .Cos\beta -Sin\alpha .Sin\beta$..... (2)

=$(1-Cos2\beta)+4Sin\alpha ,Cos\alpha .Sin\beta .Cos\beta -4Sin^{2}\alpha.Sin^{2}\beta  +Cos(2\alpha+2\beta)$

=$(1-Cos2\beta)+Sin2\alpha  Sin2\beta -(1-Cos2\alpha ).(1-Cos2\beta)+Cos(2\alpha+2\beta)$

We have applied the formula (1) and

$Sin2\alpha =2Sin\alpha Cos\alpha$ ....(3)

=$1-Cos2\beta +Sin2\alpha .Sin2\beta -1-Cos2\alpha. Cos2\beta +Cos2\alpha +Cos2\beta +Cos2\alpha .Cos2\beta -Sin2\alpha .Sin2\beta$

Here we applied formula (2).

=$Cos2\alpha$

(Answer)

Answer 2

Answer:

$2 \sin ^{2} \beta+4 \cos (\alpha+\beta) \sin \alpha \sin \beta+\cos 2(\alpha+\beta)=\cos 2 \alpha$

Given Data:  

$2 \sin ^{2} \beta+4 \cos (\alpha+\beta) \sin \alpha \sin \beta+\cos 2(\alpha+\beta)$

Step 1:

= $2 \sin ^{2} \beta+4 \cos (\alpha+\beta) \sin \alpha \sin \beta+\cos 2(\alpha+\beta)$

We know that $\cos 2 \theta=\cos ^{2} \theta-1$

$\theta=\alpha+\beta$

Step 2:

$2 \sin ^{2} \beta+4 \cos (\alpha+\beta) \sin \alpha \sin \beta+\cos ^{2}(\alpha+\beta)-1$

$2 \sin ^{2} \beta-1+2 \cos (\alpha+\beta)(2 \sin \alpha \sin \beta+\cos (\alpha+\beta))$

$-\left(1,1-2 \sin ^{2} \beta\right)+2 \cos (\alpha+\beta)(2 \sin \alpha \sin \beta+2 \cos \alpha \cos \beta)-\sin \alpha \sin \beta$

$-\cos 2 \beta+2 \cos (\alpha+\beta) \cos (\alpha-\beta)$

Step 3:

We know that 2 \cos A \cos B=\cos (A+B)+\cos (A-B)  (Apply this formulae in the below expression)

$=-\cos 2 \beta+\cos (\alpha+\beta+\alpha-\beta)+\cos (\alpha+\beta-\alpha+\beta)$………….. (-α and+α gets cancelled)

$=-\cos 2 \beta+\cos 2 \alpha+\cos 2 \beta$

Step 4:

$2 \sin ^{2} \beta+4 \cos (\alpha+\beta) \sin \alpha \sin \beta+\cos 2(\alpha+\beta)$

$=\cos 2 \alpha$

Hence Proved.