MCQ
If $A + B + C = \frac{{3\pi }}{2},$ then $\cos 2A + \cos 2B + \cos 2C = $
- A$1 - 4\cos A\,\cos B\,\cos C$
- B$4\sin A\,\,\sin B\,\,\sin C$
- C$1 + 2\cos A\,\cos B\,\cos C$
- ✓$1 - 4\sin A\,\,\sin B\,\,\sin C$
✓
Answer
Correct option: D.
$1 - 4\sin A\,\,\sin B\,\,\sin C$
d
(d) $\cos 2A + \cos 2B + \cos 2C$
(d) $\cos 2A + \cos 2B + \cos 2C$
$ = 2\cos (A + B)\cos (A - B) + \cos 2C$
$ = 2\cos \left( {\frac{{3\pi }}{2} - C} \right)\cos (A - B) + \cos 2C$
$ = - 2\sin C\cos (A - B) + 1 - 2{\sin ^2}C$
$ = 1 - 2\sin C\{ \cos (A - B) + \sin C\} $
$ = 1 - 2\sin C\left\{ {\cos (A - B) + \sin \left( {\frac{{3\pi }}{2} - (A + B)} \right)} \right\}$
$ = 1 - 2\sin C\{ \cos (A - B) - \cos (A + B)\} $
$ = 1 - 4\sin A\sin B\sin C$.
Trick : Check by assuming $A = B = C = \frac{\pi }{2}$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$\mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{{x^2} + 5x + 3}}{{{x^2} + x + 3}}} \right)^x}$=
→The area of the triangle whose vertices are complex numbers $z,\, iz,\, z + iz$ in the Argand diagram is
→If $\lim _{x \rightarrow 0}\left[1+x \ln \left(1+b^2\right)\right]^{\frac{1}{x}}=2 b \sin ^2 \theta, b>0 \text { and } \theta \in(-\pi, \pi],$ then the value of $\theta$ is
→If the straight line $ax + by + c = 0$ always passes through $(1, -2),$ then $a, b, c$ are
→The angle between the lines $x\cos {\alpha _1} + y\sin {\alpha _1} = {p_1}$ and $x\cos {\alpha _2} + y\sin {\alpha _2} = {p_2}$is
→If the equation of the locus of a point equidistant from the points $({a_1},{b_1})$ and $({a_2},{b_2})$ is $({a_1} - {a_2})x + ({b_1} - {b_2})y + c = 0$, then the value of $c$ is
→The feasible solution for a LPP is shown in Figure Let $z=3 x-4 y$ be the objective function. (Maximum value of $z+$ Minimum value of $z$ ) is equal to $....$
→Let $f(x) = x - [x],$ for every real number $x,$ where $[x]$ is the integral part of $x.$ Then $\int_{ - 1}^1 {f(x)\,dx} $=
→The solution of equation $\frac{{p + q - x}}{r} + \frac{{q + r - x}}{p}$+$\frac{{r + p - x}}{q}$+$\frac{{3x}}{{p + q + r}} = 0$ is
→For $n \in N$, let $S _{ n }=\left\{ z \in C :| z -3+2 i |=\frac{ n }{4}\right\}$ and $T _{ n }=\left\{ z \in C :| z -2+3 i |=\frac{1}{ n }\right\}$ Then the number of elements in the set $\left\{ n \in N : S _{ a } \cap T _{ n }=\phi\right\}$ is.
→