# Need Help with Trigonometry Problems

#### tom3217

##### New Member
I'm studying for Trigonometry exam but there are fews problems that involves Trig Identities that i don't know how to solve. Please help me solve these problems asap.
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a) 1 - sin x = cos^2x
b) cosx cos2x + 1/2 = sin2x sinx
c) cos x = sin2x
d) If a and b are second-quadrant angles such that sin a= 4/5 and tan= -12/5, find sin(a+b).
my issue has been solved

Last edited:

#### anemone

##### Paris la ville de l'amour
Staff member
Moderator
Math Helper
For the first problem, can you rewrite the right hand side of the expression, i.e. $$\displaystyle \cos^2 x$$ in terms of $$\displaystyle \sin x$$ function? Once you have done that, you then have a quadratic equation in $$\displaystyle \sin x$$ and you can use the quadratic formula or factoring method to solve for x.

MarkFL

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
For the second problem, I would use these two product-to-sum identities:

$$\displaystyle \cos(\alpha)\cos(\beta)=\frac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}{2}$$

$$\displaystyle \sin(\alpha)\sin(\beta)=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}$$

And so, the equation then becomes:

$$\displaystyle \frac{\cos(x)+\cos(3x)}{2}+\frac{1}{2}=\frac{\cos(x)-\cos(3x)}{2}$$

Multiply through by 2:

$$\displaystyle \cos(x)+\cos(3x)+1=\cos(x)-\cos(3x)$$

Arrange as:

$$\displaystyle \cos(3x)=-\frac{1}{2}$$

And so we know:

$$\displaystyle 3x=\pi\pm\frac{\pi}{3}+2\pi k=\frac{\pi}{3}(3\pm1+6k)$$ where $$k\in\mathbb{Z}$$

Hence:

$$\displaystyle x=\frac{\pi}{9}(3\pm1+6k)$$

Does that make sense?

anemone

Staff member