- #1

chwala

Gold Member

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- 274

- Homework Statement:
- see attached

- Relevant Equations:
- integration

Find question here,

My approach, using cosine sum and product concept, we shall have;

##\cos (A+B)-\cos (A-B)=-2\sin A\sin B##

##⇒\cos D-\cos C=-2\sin\dfrac{C+D}{2} \sin\dfrac {C-D}{-2}##

##⇒-3[\cos(A+B)-\cos(A-B)]=6\sin A sinB##

We are given ##A=4θ## and ##B=2θ##, therefore,

##⇒-3[\cos 6θ-\cos 2θ]=6\sin 4θ\sin2θ##

##⇒3\cos 2θ-3\cos 6θ=6\sin 4θ\sin2θ##

I made use of chain rule in my working... i.e letting ##u=2θ## and ##u=6θ## ...

##\int_0^\frac{π}{3} [3\cos 2θ-3\cos 6θ+2] dθ=\dfrac{3}{2} \sin 2θ-\dfrac{1}{2} \sin 6θ +2θ## from ##[θ=0]## to ##[θ=\frac{π}{3}]##

On substituting the given limits, we end up with,

##[1.2990-0+2.094395]-[0]=3.393## to 3 decimal places...

Any other approach is welcome...cheers

My approach, using cosine sum and product concept, we shall have;

##\cos (A+B)-\cos (A-B)=-2\sin A\sin B##

##⇒\cos D-\cos C=-2\sin\dfrac{C+D}{2} \sin\dfrac {C-D}{-2}##

##⇒-3[\cos(A+B)-\cos(A-B)]=6\sin A sinB##

We are given ##A=4θ## and ##B=2θ##, therefore,

##⇒-3[\cos 6θ-\cos 2θ]=6\sin 4θ\sin2θ##

##⇒3\cos 2θ-3\cos 6θ=6\sin 4θ\sin2θ##

I made use of chain rule in my working... i.e letting ##u=2θ## and ##u=6θ## ...

##\int_0^\frac{π}{3} [3\cos 2θ-3\cos 6θ+2] dθ=\dfrac{3}{2} \sin 2θ-\dfrac{1}{2} \sin 6θ +2θ## from ##[θ=0]## to ##[θ=\frac{π}{3}]##

On substituting the given limits, we end up with,

##[1.2990-0+2.094395]-[0]=3.393## to 3 decimal places...

Any other approach is welcome...cheers

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