If $\cos ^2 48^{\circ}-\sin ^2 12^{\circ}=$

$\frac{\sqrt{3}+1}{2 \sqrt{2}}$

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

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Application +4 | -1

\cos ^2 48^{\circ}-\sin ^2 12^{\circ}=

View answer and explanation

Correct option: B

\frac{\sqrt{5}+1}{8}

Option B (

\frac{{\sqrt{{5}}+1}}{{8}}

) is correct because it is derived from using the identity

\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B)

and the value of

\cos 36^{\circ}

Practice All Chapter MCQs →

\tan \alpha=\frac{p}{q}

\alpha=6 \beta

\alpha

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

\triangle A B C

C

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=