$cos60^@ xx cos30^@+sin60^@ xx sin30^@$ ?

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Answer 1 · 2018-03-18T07:40:58

The value is $sqrt3/2$ .

You can do each calculation separately and do all the math, or you can recognize the angle difference formula for cosine:

$cos(A-B)=cosAxxcosB+sinAxxsinB$

In this case, $A$ is $60^@$ and $B$ is $30^@$ . Use the formula backward now:

$color(white)=cos60^@ xx cos30^@+sin60^@ xx sin30^@$

$=cos(60^@-30^@)$

$=cos(30^@)$

$=sqrt3/2$

We can verify our answer by doing all the calculations:

$color(white)=cos60^@ xx cos30^@+sin60^@ xx sin30^@$

$=1/2xxsqrt3/2+sqrt3/2xx1/2$

$=sqrt3/4+sqrt3/4$

$=(2sqrt3)/4$

$=sqrt3/2$

Our answer is correct. Hope this helped!

Answer 2 · 2018-03-18T08:09:31

$Cos60^@×cos30^@+sin60^@×sin30^@$

This is in the form,
$cos A cos B + sin A sin B = cos(A - B)$

$=> cos(60^@-30^@)$
$=> cos30^@$
$=> sqrt3/2$

Alternatively ,

$Cos60^@×cos30^@+sin60^@×sin30^@$

Plug in the values from the standard trig table.

$=>1/2*sqrt3/2 + sqrt3/2*1/2$
$=> sqrt3/4 + sqrt3/4$
$=>(cancel2sqrt3)/cancel4^2$
$=> sqrt3/2$

cos60^@ xx cos30^@+sin60^@ xx sin30^@cos60^@ xx cos30^@+sin60^@ xx sin30^@?