Question

The value of cos 78degrees cos 12 degrees - sin 27 degrees cos 3 degrees is​

Answer 1

Please find the attachment

Answer 2

Answer:

Using trigonometric formula of $cosA cosB$ and $sinA cosB$ and we can find the value of $cos78\textdegree cos 12\textdegree-sin27\textdegree cos3\textdegree$.

Final Answer is: $cos78\textdegree cos 12\textdegree-sin27\textdegree cos3\textdegree=-\frac{1}{4}$

Step-by-step explanation:

Our question in the form of  $cosA cosB- sinAcosB$.

So that we should the trigonometric formula for $cosA cosB$ and $sinA cosB$.

Already we know the formula,

                  $2cosA cosB= cos(A+B)+cos(A-B)$

We need $cosA cosB$ only. So that,

                    $cosA cosB=\frac{1}{2} [cos(A+B)+cos(A-B)]$

Now we can solve $cos78\textdegree cos 12\textdegree$ using the above formula. Here $A=78\textdegree$ and $B=12\textdegree$

               $cos78\textdegree cos12\textdegree=\frac{1}{2} [cos(78\textdegree+12\textdegree)+cos(78\textdegree-12\textdegree)]$  

                                     $=\frac{1}{2} [cos90\textdegree+cos66\textdegree]$

Using trigonometric table, $cos 90\textdegree=0$.

                                     $=\frac{1}{2} [0+cos66\textdegree]$

                                     $=\frac{1}{2} cos66\textdegree$

and $cos 66\textdegree$ can be written as, $cos(90\textdegree - 24\textdegree)$.

                                     $=\frac{1}{2} cos(90\textdegree-24\textdegree)$

Trigonometric formula of $cos(90\textdegree - \theta) = sin \theta$. That is $cos(90\textdegree-24\textdegree) = sin 24\textdegree$

                $cos78\textdegree cos12\textdegree=\frac{1}{2} sin24\textdegree$  ------------------(1)

Similarly, next we will find $sin27\textdegree cos3\textdegree$. For that we use the following formula.

                    $2sinA cosB= sin(A+B)+sin(A-B)$

we need $sinAcosB$ only. So that,

                      $sinA cosB=\frac{1}{2} [ sin(A+B)+sin(A-B)]$

Now we can solve $sin27\textdegree cos3\textdegree$ using the above formula. Here $A=27\textdegree$ and $B=3\textdegree$

                   $sin27\textdegree cos3\textdegree=\frac{1}{2} [ sin(27\textdegree +3\textdegree)+sin(27\textdegree -3\textdegree)]$

                                       $=\frac{1}{2} [ sin30\textdegree+sin24\textdegree]$

Using trigonometric table, $sin30\textdegree=\frac{1}{2}$.

                    $sin27\textdegree cos3\textdegree=\frac{1}{2} [ \frac{1}{2} +sin24\textdegree]$ ------------(2)

Now subtract equation(1) and equation(2).

$cos78\textdegree cos 12\textdegree-sin27\textdegree cos3\textdegree=\frac{1}{2} sin 24\textdegree-\frac{1}{2} (\frac{1}{2} +sin24\textdegree)$

                                             $=\frac{1}{2} [sin 24\textdegree-( \frac{1}{2} +sin24\textdegree)]$

                                             $=\frac{1}{2} [sin 24\textdegree- \frac{1}{2} -sin24\textdegree]$

                                             $=\frac{1}{2} (- \frac{1}{2} )$

 $cos78\textdegree cos 12\textdegree-sin27\textdegree cos3\textdegree=-\frac{1}{4}$