Question
Question: Show that: \[\cos \left( {{{35}^ \circ } + A} \right) \cdot \cos \left( {{{35}^ \circ } - B} \right)...
Show that: cos(35∘+A)⋅cos(35∘−B)+sin(35∘+A)⋅sin(35∘−B)=cos(A+B).
Solution
Here in this question, we have to prove the given trigonometric function by showing the left hand side is equal to the right hand side (i.e., L.H.S=R.H.S). To solve this, we have to consider L.H.S and simplify by using a formula of cosine and sum identity and by further simplification we get the required solution.
Complete step by step solution:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function.
Consider the given question:
show that
⇒cos(35∘+A)⋅cos(35∘−B)+sin(35∘+A)⋅sin(35∘−B)=cos(A+B)--------(1)
Consider Left hand side of equation (1)
⇒L.H.S
⇒cos(35∘+A)⋅cos(35∘−B)+sin(35∘+A)⋅sin(35∘−B)--------(3)
Now, expand each term using a trigonometric formula of sum and difference identity i.e.,
Sine sum identity: sin(A+B)=sinA⋅cosB+cosA⋅sinB
Sine difference identity: sin(A−B)=sinA⋅cosB−cosA⋅sinB
Cosine sum identity: cos(A+B)=cosA⋅cosB−sinA⋅sinB
Cosine difference identity: cos(A−B)=cosA⋅cosB+sinA⋅sinB
On substituting the formulas the equation (2) becomes
⇒(cos(35∘)⋅cos(A)−sin(35∘)⋅sin(A))⋅(cos(35∘)⋅cos(B)+sin(35∘)⋅sin(B)) +(sin(35∘)⋅cos(A)+cos(35∘)⋅sin(A))⋅(sin(35∘)⋅cos(B)−cos(35∘)⋅sin(B))
On multiplication, we have
⇒cos2(35∘)⋅cos(A)cos(B)+cos(35∘)sin(35∘)⋅cos(A)sin(B) −cos(35∘)sin(35∘)⋅sin(A)cos(B)−sin2(35∘)⋅sin(A)sin(B) +sin(35∘)2⋅cos(A)cos(B)−sin(35∘)cos(35∘)⋅cos(A)sin(B) +cos(35∘)sin(35∘)⋅sin(A)cos(B)−cos2(35∘)⋅sin(A)sin(B)
on simplification and rearranging, we have
⇒cos2(35∘)⋅cos(A)cos(B)−cos2(35∘)⋅sin(A)sin(B) +sin(35∘)2⋅cos(A)cos(B)−sin2(35∘)⋅sin(A)sin(B)
Take out common terms, then
⇒cos2(35∘)⋅(cos(A)cos(B)−sin(A)sin(B))+sin(35∘)2(cos(A)cos(B)−sin(A)sin(B))
⇒(cos(A)cos(B)−sin(A)sin(B))(cos2(35∘)+sin(35∘)2)
As we now the trigonometric identity i.e., cos2θ+sin2θ=1, then
⇒(cos(A)cos(B)−sin(A)sin(B))(1)
by using cosine sum identity, the above equation becomes
⇒cos(A+B)
⇒R.H.S
Therefore, L.H.S=R.H.S
cos(35∘+A)⋅cos(35∘−B)+sin(35∘+A)⋅sin(35∘−B)=cos(A+B)
Hence proved.
Note:
When solving trigonometry based questions, we have to know the definitions of ratios and always remember the standard angles and formulas are useful for solving certain integration problems where a cosine and sum identity may make things much simpler to solve. Thus, in math as well as in physics, these formulae are useful to derive many important identities.