Question

If $A+B+C={270}^o$ then $\cos 2A+\cos 2B+\cos 2C$=

• A. $1-4\sin A\sin B\sin C$
• B. $1-4\cos A\cos B\cos C$
• C. $4\sin A\sin B\sin C$
• D. $1+2\cos A\cos B\cos C$

Answer 1

The correct option is A $1-4\sin A\sin B\sin C$

$A+B+C={270}^{\circ }$

them $cos2a+cos2b+cos2c...\left(1\right)$

Let take first 2 terms,

$cos2a+cos2b=2cos(a+b)cos(a-b)$

$=2cos(270-c)cos(a-b)$

$=-2sinccos(a-b)$

Now $cos2c=1-2si{n}^{2}c$

Using (1),

$cos2a+cos2b+cos2c$

$=-2sinccos(a-b)+cos2c$

$-2sinccos(a-b)+si{n}^{2}2c$

$\Rightarrow 1-2sinnc\left(cos\right(a-b)+si{n}^{2}c)$

$[\because cos(a-b)+sin\{\frac{3x}{2}-(a+b\left)\right\}]$

$=1-2sinc\left[cos\right(a-b)-cos(a+b\left)\right]$

$=1-2sinc\left(2sinasinb\right)$

$=1-4sinasinbsinc$

So, option (a) is correct