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C3 Trig

Hi, i ma having trouble with the questions below, if anyone could help that would be really useful. Is there an easy way to do them.....i find them really tricky! (i especially struggle on the proofs!):confused:

1. Prove that tanθ +cotθ = 2 cosec 2θ

2. Solve by getting the exact answers in terms of π, 2(1- cos 2&#952:wink: = tan θ

3. Express sin^2θ in the form a + b cos 2θ, where a and b are constants.

4. Solve the equation 3 3 cos 2&#952; = 2 cosec &#952;, giving all solutions in radians in the interval 0 < &#952; < 2&#960;

Thanks
Reply 1
1. think about sin^2+cos^2 = 1 and sin2x=2sinxcosx... that made it possible for me to solve it...
Reply 2
Q1:

RHS2csc2θ\displaystyle\huge \text{RHS} \equiv 2\csc 2\theta
2sin2θ\displaystyle\huge \equiv \frac{2}{\sin 2\theta}
22sinθcosθ\displaystyle\huge \equiv \frac{2}{2\sin \theta \cos \theta}
1sinθcosθ\displaystyle\huge \equiv \frac{1}{\sin \theta \cos \theta}
sin2θ+cos2θsinθcosθ\displaystyle\huge \equiv \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}
sin2θsinθcosθ+cos2θsinθcosθ\displaystyle\huge \equiv \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta}
sinθcosθ+cosθsinθ\displaystyle\huge \equiv \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}
tanθ+cotθLHS\displaystyle\huge \equiv \tan \theta + \cot \theta \equiv \text{LHS}

TSR's LaTeX system is crap... Those silly floating twos are supposed to be exponents, but for some reason decided they'd be happier riding on top of their trig functions instead of squaring them. :rolleyes:
Reply 3
Dez
Q1:TSR's LaTeX system is crap... Those silly floating twos are supposed to be exponents, but for some reason decided they'd be happier riding on top of their trig functions instead of squaring them. :rolleyes:


Thanks a lot
Reply 4
nota bene
1. think about sin^2+cos^2 = 1 and sin2x=2sinxcosx... that made it possible for me to solve it...


thanks for the help
Reply 5
3. Express sin^2&#952; in the form a + b cos 2&#952;, where a and b are constants.


cos 2&#952; = 1 - 2sin²&#952;

Re-arranging:

2sin²&#952; = 1 - cos2&#952;

sin²&#952; = 1/2 - (cos2&#952:wink:/2

So a = 1/2, b = -1/2.

Hope this helps,

~~Simba
Reply 6
Simba
cos 2&#952; = 1 - 2sin²&#952;

Re-arranging:

2sin²&#952; = 1 - cos2&#952;

sin²&#952; = 1/2 - (cos2&#952:wink:/2

So a = 1/2, b = -1/2.

Hope this helps,

~~Simba


thanks for that - just looking at you signature - that is amazing getting 100 in all modules :biggrin:
Reply 7
4. Solve the equation 3 &#8211; 3 cos 2&#952; = 2 cosec &#952;, giving all solutions in radians in the interval 0 < &#952; < 2&#960;

Unparseable latex formula:

\tex \large 3-3cos 2\theta = 2 cosec \theta



Unparseable latex formula:

\tex \large 3 - 3(1-2sin^2\theta) = \frac{2}{sin\theta}



Unparseable latex formula:

\tex \large 6sin^2\theta = \frac{2}{sin\theta}



Unparseable latex formula:

\tex \large sin^3\theta = \frac{1}{3}



Unparseable latex formula:

\tex \large sin \theta = \frac{1}{^3\sqrt{3}}



Unparseable latex formula:

\tex \large \theta = 0.77, 2.38, 3.91, 5.32



The root 3 is the cubic root of 3 not 3root3

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