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If cos(B/3) = 1/2(m+1/m) then, prove that: cosB = 1/2(m^3 + 1/m^3)

can't even start pls hlp
ramu asked 2 years ago·

Solution:

Given: cosB3=12(m+1m)\cos\frac{B}{3} = \frac{1}{2}\left(m+\frac{1}{m}\right)

To prove: cosB=12(m3+1m3)\cos B = \frac{1}{2}\left(m^3 + \frac{1}{m^3}\right)

We'll use the triple angle formula for cosine:

cos(3θ)=4cos3θ3cosθ\cos(3\theta) = 4\cos^3\theta - 3\cos\theta

Let θ=B3\theta = \frac{B}{3}, then:

cosB=cos(3×B3)=4cos3B33cosB3\cos B = \cos\left(3 \times \frac{B}{3}\right) = 4\cos^3\frac{B}{3} - 3\cos\frac{B}{3}

Substituting the given value cosB3=12(m+1m)\cos\frac{B}{3} = \frac{1}{2}\left(m+\frac{1}{m}\right):

cosB=4×[12(m+1m)]33×12(m+1m)\cos B = 4 \times \left[\frac{1}{2}\left(m+\frac{1}{m}\right)\right]^3 - 3 \times \frac{1}{2}\left(m+\frac{1}{m}\right)

=4×18(m+1m)332(m+1m)= 4 \times \frac{1}{8}\left(m+\frac{1}{m}\right)^3 - \frac{3}{2}\left(m+\frac{1}{m}\right)

=12(m+1m)332(m+1m)= \frac{1}{2}\left(m+\frac{1}{m}\right)^3 - \frac{3}{2}\left(m+\frac{1}{m}\right)

Now, let's expand (m+1m)3\left(m+\frac{1}{m}\right)^3:

(m+1m)3=m3+3m2×1m+3m×1m2+1m3\left(m+\frac{1}{m}\right)^3 = m^3 + 3m^2 \times \frac{1}{m} + 3m \times \frac{1}{m^2} + \frac{1}{m^3}

=m3+3m+3m+1m3= m^3 + 3m + \frac{3}{m} + \frac{1}{m^3}

=m3+1m3+3(m+1m)= m^3 + \frac{1}{m^3} + 3\left(m + \frac{1}{m}\right)

Substituting back:

cosB=12[m3+1m3+3(m+1m)]32(m+1m)\cos B = \frac{1}{2}\left[m^3 + \frac{1}{m^3} + 3\left(m + \frac{1}{m}\right)\right] - \frac{3}{2}\left(m+\frac{1}{m}\right)

=12(m3+1m3)+32(m+1m)32(m+1m)= \frac{1}{2}\left(m^3 + \frac{1}{m^3}\right) + \frac{3}{2}\left(m + \frac{1}{m}\right) - \frac{3}{2}\left(m+\frac{1}{m}\right)

=12(m3+1m3)+32(m+1m)32(m+1m)= \frac{1}{2}\left(m^3 + \frac{1}{m^3}\right) + \frac{3}{2}\left(m + \frac{1}{m}\right) - \frac{3}{2}\left(m + \frac{1}{m}\right)

=12(m3+1m3)= \frac{1}{2}\left(m^3 + \frac{1}{m^3}\right)

Hence proved.

dibas answered 25 days ago