The identity has two equal side
Let us check each one
A)
[tex]\sin 8x=2\sin 4x\cos 4x[/tex]The L.H.S is sin 8x, we will use the rule of the double angle
[tex]\sin 2\theta=2\sin \theta\cos \theta[/tex]Then divide 8x by 2 and use the rule of the double angle
[tex]\sin 8x=2\sin 4x\cos 4x[/tex]This value = the R.H.S, then
sin 8x = 2sin 4x cos 4x is an identity
Then answer A is an identity
C)
[tex](\sin x-\cos x)^2=1+2\sin 2x[/tex]Solve the left side
[tex]\begin{gathered} LHS=(\sin x-\cos x)^2 \\ LHS=\sin ^2x+\cos ^2x-2\sin x\cos x \end{gathered}[/tex]Since sin^2(x) + cos^2(x) = 1, then
[tex]LHS=1-2\sin x\cos x[/tex]Since 2 sin x cos x = sin(2x), then
[tex]LHS=1-\sin 2x[/tex]But the R.H.S is 1 + sin 2x, then
L.H.S not equal R.H.S
Then answer C is not an identity
D)
[tex]\frac{1-\tan^2x}{2\tan x}=\frac{1}{\tan 2x}[/tex]Since tan 2x is equal to
[tex]\tan 2x=\frac{2\tan x}{1-\tan ^2x}[/tex]By reciprocal the two sides, then
[tex]\frac{1}{\tan2x}=\frac{1-\tan ^2x}{2\tan x}[/tex]Switch the 2 sides
[tex]\frac{1-\tan^2x}{2\tan x}=\frac{1}{\tan 2x}[/tex]Answer D is an identity
B)
[tex]\frac{\sin x+\sin5x}{\cos x+\cos5x}=\tan 3x[/tex]For the left side
[tex]L.H.S=\frac{\sin x+\sin 5x}{\cos x+\cos 5x}[/tex]Change angle 5x to 4x + x
[tex]\begin{gathered} \sin 5x=\sin (4x+x) \\ \sin (4x+x)=\sin 4x\cos x+\cos 4x\sin x \end{gathered}[/tex]Solve sin4x
[tex]\sin 4x\cos x=(2\sin 2x\cos 2x)(\cos x)=2\cos x\sin 2x\cos 2x[/tex]Solve cos4x
[tex]\cos 4x\sin x=(1-2\sin ^22x)(\sin x)=\sin x-2\sin x\sin ^22x[/tex]Then sinx+
