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3.Trigonometrical Ratios, Functions and Identities
easy
સાબિત કરો કે : $\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x=\cos x$
Option A
Option B
Option C
Option D
Solution
$L.H.S.$ $=\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x$
$=\frac{1}{2}[2 \sin (n+1) x \sin (n+2) x+2 \cos (n+1) x \cos (n+2) x]$
$=\frac{1}{2}\left[\begin{array}{c}\cos \{(n+1) x-(n+2) x\}-c i s\{(n+1) x+(n+2) x\} \\ +\cos \{(n+1) x+(n+2) x\}+\cos \{(n+1) x-(n+2) x\}\end{array}\right]$
$\left[\begin{array}{c}\because-2 \sin A \sin B=\cos (A+B)-\cos (A-B) \\ 2 \cos A \cos B=\cos (A+B)+\cos (A-B)\end{array}\right]$
$=\frac{1}{2} \times 2 \cos \{(n+1) x-(n+2) x\}$
$=\cos (-x)=\cos x= R . H.S$
Standard 11
Mathematics
