Question

cos³340+cos³390+cos³450

please give correct answer​

Answer 1

⇒cos

3

40

o

−3sin50

o

⇒cos

3

40

o

−3sin(90−40)

⇒cos

3

40−3cos40

Let cos40=cosx

by complementing trigonometric function

We can write cos(n

2

π

±θ)=cosθ

Now if cosx=cos40

then cos3x=cos120

Now ⇒cos3x=cos(2x+x)=cos2xcosx−sin2xsinx

⇒cos120=(cos

2

x−sin

2

x)cosx−2sin

2

xcosx

⇒

2

−1

=cos

3

x−(1−cos

2

x)cosx−2sin

2

xcosx

⇒

2

−1

=cos

3

x−cosx+cos

3

x−2(1−cos

2

x)cosx

⇒

2

−1

=2cos

3

x−cosx−2cosx+2cos

3

x

⇒8cos

3

x−6cosx+1=0

Solving this we get

cosx=−0.94,0.174.0.766

as cosx cannot be negative between [0,

2

π

] and always decreasing i.e cos45<cos40 and cos45=0.707

∴cosx=cos40=0.766

∴cos

3

40−3cos40=−1.848

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