Show that cos 2-cos 2- - sin 2- - - - sin 2- - - - cos 2- - 2-

LHS=cos 2θ #160; cos 2ϕ #160; +sin ^ 2 (θ ϕ ) sin ^ 2 (θ +ϕ ) =cos ^ 2 θ #160; cos 2ϕ #160; sin (θ ϕ +θ +ϕ ) sin (θ ϕ θ ϕ ) = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trigonometric Equations

Show that co...

Question

Show that
$cos 2 θ cos 2 ϕ + {sin}^{2} (θ - ϕ) - {sin}^{2} (θ + ϕ) = cos (2 θ + 2 ϕ)$

Open in App

Solution

Suggest Corrections

Similar questions

cos 2 θ cos 2 ϕ + {sin}^{2} (θ - ϕ) - {sin}^{2} (θ + ϕ) = cos (2 θ + 2 ϕ)

cos 2 θ cos 2 ϕ + {sin}^{2} (θ - ϕ) - {sin}^{2} (θ + ϕ) = cos 2 (θ + ϕ)

If $s i n 2 θ + s i n 2 ϕ = \frac{1}{2}$ and $c o s 2 θ + c o s 2 ϕ = \frac{3}{2}$ , then $c o s^{2} (θ - ϕ) =$

\frac{1}{2}

\frac{3}{2}

\frac{3}{8}

\frac{5}{8}

\frac{3}{4}

\frac{5}{4}

(a + b) t a n (θ - ϕ) = (a - b) t a n (θ + ϕ)

\frac{s i n (2 θ)}{s i n (2 ϕ)}

Join BYJU'S Learning Program