value of an inverse trignometric expression

Question

How can we find the value of $ 3\sin(\frac12\arccos\frac19)+ 4\cos(\frac12\arccos\frac18)$ ? Substituting A = $\arccos\frac19$

My approach to this question.. I tried to use the formula $\cos A = \cos^2\frac A2\ - \sin^2\frac A2 $ but it led to a negative value under radical.. Any hints?

Answer 1

Hint:

since $A$ is in the first quadrant, using the half-angle formulas $$ \sin (A/2)=\sqrt{\frac{1-\cos A}{2}} \qquad \cos (A/2)=\sqrt{\frac{1+\cos A}{2}} $$

your expression becomes: $$ 3\sqrt{\frac{1-1/9}{2}}+4\sqrt{\frac{1+1/8}{2}} $$

can you do from this?

Answer 2

Use the formulae: $1-\cos A=2\sin^2\frac{A}{2},1+\cos x=2\cos^2\frac{A}{2}$.

So $\frac{8}{9}=2\sin^2\frac{1}{2}A$ and hence $\sin\frac{1}{2}A=\frac{2}{3}$ (note that the angle is obviously in the first quadrant, so we want the positive sign).

Similarly, $\cos\frac{1}{2}A=\frac{\sqrt5}{3}$, so $3\sin\frac{1}{2}A+4\cos\frac{1}{2}A=2+\frac{4\sqrt5}{3}$.

Answer 3

Let $\alpha=\arccos\frac19$, $\;\beta=\arccos\frac18$. Note $0<\alpha,\,\beta<\frac\pi 2$.

• $\cos^2\frac\alpha2=\dfrac{1+\cos\alpha}{2}=\dfrac{1+\frac19}{2}=\dfrac59$.
• $\sin^2\frac\alpha2+\cos^2\frac\alpha2=1$, whence $\sin \frac\alpha2=\sqrt{1-\frac59}=\dfrac23$.
• Similarly $\cos^2\frac\beta2=\frac9{16}$, whence $\;\cos\frac\beta2=\dfrac 34$.

Finally, one gets $\;3\sin(\frac12\arccos\frac19)+ 4\cos(\frac12\arccos\frac18)=2+3=5.$

Answer 4

Hint:

You aren't so far.

Use

$$\cos(A)=2\cos^2\left(\frac A2\right)-1=1-2\sin^2\left(\frac A2\right),$$

giving

$$\cos\left(\frac A2\right)=\pm\sqrt{\frac{1+\cos(A)}2},\ \sin\left(\frac A2\right)=\pm\sqrt{\frac{1-\cos(A)}2}.$$