Posted by **Suzy** on Friday, August 16, 2013 at 9:36pm.

2. solve cos 2x-3sin x cos 2x=0 for the principal values to two decimal places.

3. solve tan^2 + tan x-1= 0 for the principal values to two decimal places.

4. Prove that tan^2(x) -1 + cos^2(x) = tan^2(x) sin^2 (x).

5.Prove that tan(x) sin(x) + cos(x)= sec(x)

6.Prove that tan(x) cos^2(x)+sin^2(x)= cos(x)+ sin(x)

7.Prove that 1+tan(x)/1-tan(x)= sec^2(x)+ 2tan(x)/1-tan^2(x)

8.Prove that sin^2(x)-cos^2(x)/tan(x)sin(x)+cos(x)tan(x)=cos(x)-con(x)cos(x)

9. find a counterexample to show that the equation sec(x)-cos(x)=sin(x) sec(x) is not an identity

- Trigonometry desperate help, clueless girl here -
**Steve**, Saturday, August 17, 2013 at 12:37am
#2

cos2x - 3sinx cos2x = 0

(cos2x)(1-3sinx) = 0

as you know, if the product of two numbers is zero, one or the other must be zero. So, cos2x = 0 or 1-3sinx = 0

cos2x=0 means x is pi/4,3pi/4,5pi/4,7pi/4

1-3sinx=0 means x = arcsin(1/3) = .3398

But, you need all angles between 0 and 2pi, so since sinx >0 in Qi and QII,

x = .3398 or pi-.3398=2.8018

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#3.

did this one already also. What was unclear?

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#4

possibly the most useful trig identity is sin^2 x + cos^2 x = 1. You have

tan^2 x - 1 + cos^2 x

tan^2 x - (1-cos^2 x)

tan^2 x - sin^2 x

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#5.

tanx sinx + cosx

sinx/cosx * sinx + cosx

(sin^2 x + cos^2 x)/cosx

1/cosx

secx

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#6.

Must be a typo. If x=pi/4,

tan(x) cos^2(x)+sin^2(x) = 1*1/2 + 1/2 = 1

cos(x)+sin(x) = 1/√2+1/√2 = √2

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#7.

(1+tanx)/(1-tanx)

(1+tanx)^2 / (1-tan^2 x)

(1 + 2tanx + tan^2 x)/(1-tan^2 x)

(sec^2 x + 2tanx)/(1-tan^2 x)

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#8.

(sin^2 x - cos^2 x)/(tanx*sinx + cosx*tanx)

(sinx-cosx)(sinx+cosx)/(tanx(sinx+cosx))

(sinx-cosx)/tanx

sinx*cotx - cosx*cotx

cosx - cosx*cotx

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#9.

Usually a familiar angle will do the trick. If x=pi/4,

secx-cosx = √2 - 1/√2

sinx*secx = 1/√2*√2 = 1

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