Prove that-sin 2 - 8-x 2-sin 2 - 8-x 2-12 sin x

LHS=sin2 #960;8+x2 sin2 #960;8 x2 #160; #160; #160; #160; #160; #160;=121 cos2 #960;8+x2 121 cos2 #960;8 x2 #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Monotonicity in an Interval

Prove that:si...

Question

Prove that:
$\sin^{2} (\frac{π}{8} + \frac{x}{2}) - \sin^{2} (\frac{π}{8} - \frac{x}{2}) = \frac{1}{\sqrt{2}} \sin x$

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Solution

$LHS = \sin^{2} (\frac{π}{8} + \frac{x}{2}) - \sin^{2} (\frac{π}{8} - \frac{x}{2}) = \frac{1}{2} \{1 - \cos 2 (\frac{π}{8} + \frac{x}{2})\} - \frac{1}{2} \{1 - \cos 2 (\frac{π}{8} - \frac{x}{2})\} = \frac{1}{2} \{\cos (\frac{π}{4} - x) - \cos (\frac{π}{4} + x)\}$

Using the identity $\cos C - \cos D = - 2 \sin \frac{C + D}{2} \sin \frac{C - D}{2}$ , we get

$= \frac{1}{2} \{- 2 \sin (\frac{(\frac{π}{4} - x) + (\frac{π}{4} + x)}{2}) \sin (\frac{(\frac{π}{4} - x) - (\frac{π}{4} + x)}{2})\} = - \sin \frac{π}{4} \sin (- x) = \sin \frac{π}{4} \sin x [∵ \sin (- x) = - \sin x] = \frac{1}{\sqrt{2}} \sin x = RHS Hence proved .$

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Similar questions

Q. Prove that:

\sin^{2} (\frac{π}{8} + \frac{A}{2}) - \sin^{2} (\frac{π}{8} - \frac{A}{2}) = \frac{1}{\sqrt{2}} \sin A

Q. Prove that:

\sin^{2} \frac{π}{8} + \sin^{2} \frac{3 π}{8} + \sin^{2} \frac{5 π}{8} + \sin^{2} \frac{7 π}{8} = 2

Q. Prove:

{(x + \frac{1}{x})}^{2} = x^{2} + \frac{1}{x^{2}} + 2

Q. Solve the following equations for x:

(i)

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

(ii)

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

(iii)

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

(iv) 2

\tan^{- 1}

(

\sin

x

)

=

\tan

^{$-$ 1} (2

\sin

x

x \neq \frac{π}{2}

.

(v)

\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3}

(vi)

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

Prove that $t a n^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) = \frac{π}{4} + \frac{1}{2} c o s^{- 1} x^{2} .$

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Prove that: sin2 π8+x2-sin2 π8-x2=12 sin x

LHS=sin2π8+x2-sin2π8-x2 =121-cos2π8+x2-121-cos2π8-x2 =12cosπ4-x-cosπ4+x Using the identity cosC-cosD=-2sinC+D2sinC-D2, we get =12-2sinπ4-x+π4+x2 sinπ4-x-π4+x2 =-sinπ4sin-x =sinπ4sinx ∵sin-x= -sinx =12sinx=RHSHence proved.

Prove that:
$\sin^{2} (\frac{π}{8} + \frac{x}{2}) - \sin^{2} (\frac{π}{8} - \frac{x}{2}) = \frac{1}{\sqrt{2}} \sin x$