Trigonometric Identities: Given that $2\cos(3a)=\cos(a)$ find $\cos(2a)$

Question

Given that $2\cos(3a)=\cos(a)$ find $\cos(2a)$.

$2\cos(3a)=\cos(a)$

I converted $\cos(2a)$ into $\cos^2(a)-\sin^2(a)$

Then I tried plugging in. I know this is not right, but I have no clue how to solve this. Hints please?

edit: Because I got that $\cos(2a) = 4\cos^2(3a)-1$

Answer 1

$$2\cos3\alpha+2\cos\alpha=3\cos\alpha$$ or $$4\cos2\alpha\cos\alpha=3\cos\alpha.$$ Can you end it now?

I got $\cos2\alpha=-1$ or $\cos2\alpha=\frac{3}{4}.$

I used the following identity. $$\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2},$$ which we can get by the following way. $$\cos\alpha+\cos\beta=\cos\left(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2}\right)+\cos\left(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2}\right)=$$ $$=\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}-\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}+$$ $$+\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}=$$ $$=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}.$$ We can use also the following way.

$$2\cos(2\alpha+\alpha)=\cos\alpha$$ or $$2(\cos2\alpha\cos\alpha-\sin2\alpha\sin\alpha)=\cos\alpha$$ or $$2(\cos2\alpha\cos\alpha-2\sin^2\alpha\cos\alpha)=\cos\alpha.$$ Now, if $\cos\alpha=0$ we obtain $$\cos2\alpha=-1.$$ Otherwise, we obtain $$2(\cos2\alpha-2\sin^2\alpha)=1$$ or $$2(\cos2\alpha-1+\cos2\alpha)=1,$$ which gives $$\cos2\alpha=\frac{3}{4}.$$

Answer 2

$$2\cos3a=\cos a$$ $$4\cos^23a=\cos^2a$$ $$4(1+\cos6a)=1+\cos2a$$ $$4+4(4\cos^32a-3\cos2a)=1+\cos2a$$ with $x=\cos2a$ then $$16x^3-13x+3=0$$ $$13(x^3-x)+3(x^3+1)=0$$ $$(x+1)(13x^2-13x+3x^2-3x+3)=0$$ $$(x+1)(16x^2-16x+3)=0$$ tis gives the answers $\cos2a=-1$ or $\cos2a=\dfrac14$ or $\cos2a=\dfrac34$.

Answer 3

Hint:

$$a\cos3A=b\cos A$$

$$\iff a(4\cos^3A-3\cos A)=b\cos A$$

$$\iff\cos A[4a\cos^2A-3a-b]=0$$

Use $\cos2A=2\cos^2A-1$

Answer 4

I like to split $\color{blue}{\cos(3 a) = \cos(a+2a) = \cos(a) \cos(2 a) -\sin(a) \sin(2 a)}$. This makes the expression

$$ 2 \left( \cos(a) \cos(2 a) -\sin(a) \sin(2 a) \right) = \cos(a) $$

or (re-arrange and divide by $2\cos(a)$)

$$ \frac{2 \cos(a) \cos(2 a)}{2 \cos(a)} = \frac{\cos(a)}{2 \cos(a)} + \frac{2\sin(a) \sin(2 a)}{2 \cos(a) }$$

of course $\color{blue}{ \sin(2 a) = 2\sin(a)\cos(a)}$

$$ \cos(2 a) = \tfrac{1}{2} + \frac{ 2\sin(a) \left( 2 \sin(a) \cos(a) \right)}{2 \cos(a)} = \tfrac{1}{2} + 2 \sin^2(a) $$

Now use $\color{blue}{\sin^2(a) = \tfrac{1}{2} - \tfrac{1}{2} \cos(2 a)}$ to make the above

$$ \cos(2 a) = \tfrac{1}{2} + 2 \left(\tfrac{1}{2} - \tfrac{1}{2} \cos(2 a)\right) = \tfrac{3}{2} - \cos(2 a) $$

The above is solved for

$$ \cos(2 a) = \tfrac{3}{4} $$

Answer 5

cos2a?

2cos3a = cosa

2cos3a = cosa ⟺ 2(4cosa^3-3cosa)=cosa ⟺8cosa^3 - 6cosa = cosa

8cosa^3=7cosawe divide everything by cosa

8cosa^2=7 ⟺ cosa^2=7/8

cos 2a = cos^2 - sin^2

We need to find sin^2

sin^2 + cos^2 = 1 ⟺ sin^2 = 1 - 7/8 ⟺ sin^2 = 1/8

cos 2a = 7/8 - 1/8 ⟺ cos 2a = 6/8 ⟺ cos 2a = 3/4

I'm sorry if it's not written as beautifully as above