# Trigonometry desperate help, clueless girl here

posted by .

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 -

#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
******************************
#3.
did this one already also. What was unclear?
*******************************
#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
*****************************
#5.
tanx sinx + cosx
sinx/cosx * sinx + cosx
(sin^2 x + cos^2 x)/cosx
1/cosx
secx
*****************************
#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
*****************************
#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)
******************************
#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
********************************
#9.
Usually a familiar angle will do the trick. If x=pi/4,
secx-cosx = √2 - 1/√2
sinx*secx = 1/√2*√2 = 1

### Answer This Question

 First Name: School Subject: Answer:

### Related Questions

More Related Questions

Post a New Question