# 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).
Thanks for your helps
my issue has been solved

Last edited:

#### anemone

##### Paris la ville de l'amour
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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
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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

#### MarkFL

##### La Villa Strangiato
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For the third one, apply the double angle identity for sine to the RHS...what do you then have?

In the fourth one, the tangent function has no argument...can you clarify?

anemone