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

Question

16128 views around the world

You can reuse this answer
Creative Commons License

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#