Question

prove that
sin (60+x) sin (420-x) = 1+ 2cos2x/2

Answer 1

$$\begin{align}
\sin(60+x)\sin(420-x)&=\frac{1}{2}[\cos(60+x-420+x)-\cos(60+x+420-x)]\\
&=\frac{1}2[\cos(360-2x)-\cos(480)]\\
&=\frac12\left[\cos(2x)+\frac12\right]
\end{align}$$

Answer 2

1/2[cos(60+x−420+x)−cos(60+x+420−x)]
explain me this step

Answer 3

the identitiy:

$$\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$$

Answer 4

but lhs is not cmg equal to rhs

Answer 5

you mean the identity or the application of the identity?

Answer 6

LHS is not coming equal to RHS

Answer 7

i m talking about the solution of given question by you.

Answer 8

I've adjusted the identity in application for a bit, the actual application is like,
$$\sin\left(C\right) \sin\left(D\right)=\frac{1}{2}\left[\cos \left(C -D\right)- \cos \left(C+D\right)\right]$$

Answer 9

okay , by using this identity ,now see the solution done by you

Answer 10

$$C=60+x,\quad D=420-x$$$$C-D=60+x-420+x=-(360-2x)\\ C+D=60+x+420-x=480$$

Answer 11

kk.
after putting all the values in sin (60+x) sin (420-x)

Answer 12

it's not cmg equal to = 1+ 2cos2x/2

Answer 13

Let's take an example if x=30,
$$\sin(60+30)\sin(420-30)=\sin(90)\sin(390)=\sin(90)\sin(30)=1.\frac12$$
so the answer should be 1/2 on the right side also,
taking your answer, (since it is not sure it is 1+2cos(2x)/2 or (1+2cos(2x)/2)
if its,
$$\frac{1+2\cos(2x)}{2}=\frac{1+2\cos(60)}{2}=\frac{1+2\frac12}{2}=1$$
or if its,
$$1+\frac{2\cos(2x)}{2}=1+\cos(60)=1+\frac12=\frac 32$$
so either way the answer is not correct. My derived expression gives,
$$\frac12\left(\cos(2x)+\frac12\right)=\frac12\left(\cos(60)+\frac12\right)=\frac12\left(\frac 12+\frac12\right)=\frac 12$$

So I think your expression given to be proven is not correct