Question

prove that sin(-420).cos(390) + cos(-660).sin(330) = -1 help plz ... my RHS is coming +1 ... totally confused .. help plz.. as soon as possible !!

Answer 1

sin(-420).cos(390) + cos(-660).sin(330) 
-sin(420).cos(390) + cos(660).sin(330)
-sin(360+60).cos(360+30) + cos(2.360 - 60).sin(360-30)
-sin(60). cos(30) - cos(60).sin(30)
-3/4 -1/4
-1

Answer 2

Answer:

We can prove this question using trigonometric formula.To prove that given question we should consider left hand side first.Then we can get the right hand side.

Step-by-step explanation:

From the given question,

Take L.H.S(Left Hand Side) ⇒ $sin(-420\textdegree).cos(390\textdegree) + cos(-660\textdegree).sin(330\textdegree)$

Step 1: First we should know the formula for $sin(-\theta)$ and $cos(-\theta)$.

              $sin(-\theta)=-sin\theta$

              $cos(-\theta)=cos\theta$

Step 2: Now apply the above formula in the given question.

                 L.H.S $=sin(-420\textdegree).cos(390\textdegree) + cos(-660\textdegree).sin(330\textdegree)$

• $sin(-420\textdegree)$ should be written as, $sin(-420 \textdegree)=-sin(420\textdegree)$
• $cos(-660\textdegree)$ should be written as, $cos(-660 \textdegree)=cos(660\textdegree)$

                  L.H.S $=[-sin(420\textdegree)].cos(390\textdegree) + cos(660\textdegree).sin(330\textdegree)$ --------(1)

Step 3: We know the value of $0\textdegree,30\textdegree,60\textdegree,90\textdegree,180\textdegree,360\textdegree$ only.So that our given theta value should convert into the $0\textdegree,30\textdegree,60\textdegree,90\textdegree,180\textdegree,360\textdegree$.

That is,       $420\textdegree=360\textdegree+60\textdegree$

                  $390\textdegree=360\textdegree+30\textdegree$

                  $660\textdegree=(2\times360\textdegree)-60\textdegree$

                  $330\textdegree=360\textdegree-30\textdegree$

Step 4: Now apply the above values in equation(1).

 $=[-sin(360\textdegree+60\textdegree)].cos(360\textdegree+30\textdegree) + cos[(2\times360\textdegree)-60\textdegree]).sin(360\textdegree-30\textdegree)$

Step 5: We should know the formula of following trigonometric formula

      $sin(360\textdegree+\theta)=sin\theta$

      $cos(360\textdegree+\theta)=cos\theta$

     $cos[(2\times360\textdegree)-\theta)=cos\theta$

     $sin(360\textdegree-\theta)=-sin\theta$

• Now apply the above in step 2.

  $sin(360\textdegree+60\textdegree)=sin60\textdegree$

  $cos(360\textdegree+30\textdegree)=cos30\textdegree$

$cos[(2\times360\textdegree)-60\textdegree]=cos60\textdegree$

 $sin(360\textdegree-30\textdegree)=-sin30\textdegree$

• L.H.S can be written as,

                 L.H.S $=(-sin60\textdegree).cos 30\textdegree+cos60\textdegree.(-sin30\textdegree)$

Step 6: Apply $sin60\textdegree,cos30\textdegree,sin30\textdegree,cos60\textdegree$ values in step 5 using trigonometric table.

• $sin60\textdegree=\frac{\sqrt{3} }{2}$
• $cos30\textdegree=\frac{\sqrt{3} }{2}$
• $sin30\textdegree=\frac{1 }{2}$
• $cos60\textdegree=\frac{1 }{2}$

                 L.H.S $=(-\frac{\sqrt{3} }{2} .\frac{\sqrt{3} }{2}) +(\frac{1}{2} .\frac{-1}{2} )$

                           $=(-\frac{3 }{4} ) +(-\frac{1}{4} )$   ⇒ ($\sqrt{3}\times\sqrt{3}=3$)

• Denominator is same in the above equation.So that we can add numerator.

                           $=-\frac{4 }{4}$

                           $=-1$

                 L.H.S  = R.H.S hence proved.