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# C3 trig Q watch

C3 book, ex. 7D, Q7)

a) Show that cosθ-√3sinθ can be written in the form Rcos(θ+a), with R>0 and 0<a<π/2
b) Hence sketch the graph of y=cosθ-√3sinθ, 0<a<2π, giving the coordinates of points of intersection with the axes.
2. use the expansion Rcos(θ+a) = Rcosθcosa - Rsinθsina

compare this to cosθ-√3sinθ

=> Rcosa = 1

and Rsina = √3

dividing...tana = √3

a = 60° or π/3 radians

f(θ ) = Rcos(θ + π/3)

to find R you can either do Pythagoras on 1 and √3 to give √(1² + (√3)²)

or you can rearrange one of the expressions above to get R = 1/cosa = 1/cos(π/3)

both give the value R =2

so f(θ ) = 2cos(θ + π/3)

the graph of f(θ ) is a modified version of cosθ...it has been shifted (π/3) units to the left and then stretched by a factor of 2 parallel to the y axis.

it intersects the y axis at 2cos(π/3) = 1

it intersects the x axis when (θ + π/3) = π/2 and 3π/2 and 5π/2

you just rearrange to find θ
3. I did it pretty much the same way as the bear. The only difference was that if already know by memory that cos(π/3) = ½ then you know that 2cos(π/3) = 1, giving you the value of R and a in a much simpler step. This demonstrates the importance of remembering cos and sin for important values.
4. Thanks a lot!

10a) Express 7cosθ-24sinθ in the form Rcos(θ+a), with R>0 and 0<a<90. I did this bit and got 25cos(θ+73.7), which is correct. However, I can't do the rest.
b) The graph of y=7cosθ-24sinθ meets the y axis at P. State the coordinates of P. (apparently it's meant to be 0,7 but i dont know why!)
c) Write down the maximum and minimum values of 7cosθ-24sinθ.
d) Deduce the number of solutions, in the interval 0<θ<360, of the following equations:
i) 7cosθ-24sinθ=15
ii)7cosθ-24sinθ=26
iii) 7cosθ-24sinθ=-25
5. b) It meets the Y axis where theta = 0.....hence you get 7 cos 0 - 24 sin 0, sin 0 is 0, cos 0 is 1, so the y coord is 7, the theta is 0....

c) Well.....25cos (t + 73.7) = 7 cos t - 24 sin t, so the min and max of both sides are the same. Max of cos is 1, hence the maximum value is 25, and the min is 0, hence the min is 0.

d) To do this you just set 25cos(t+73.7) = x.

i) 2 sols, cause the graph goes up thro it, and down thro it.
ii) no solutions, cause the max is 25.
iii) 1 solution because it is the min for the range.
6. Thanks! (well, i didnt really get di) and how u got that?)

Could somebody please help me with this: express 5sin²θ-3cos²θ+6sinθcosθ in the form asin2θθ+bcos2θ+c, where a, b and c are constants.

I got 2.5(1-cos²θ)-1.5(cos2θ+1)+3sin2θ so far? Is that correct/incorrect? I dont know what to do now! The answer is meant to be 1+3sin2θ-4cos2θ
7. Crap!....I got c wrong I think... very embarrassing sorry.

c SHOULD read min -25...I somehow got the impression you were only between 0 and 90 with those limits...

For d i, think of the typical cos graph. It curves down, and then back up again between 0 and 360. If you draw a line anywhere going straight across the graph(like for example at 15 like in the question), it will cross the curve at 2 points, on the way down, and the reflection on the way up.

3sin2Φ is obviously the solution to the 6sincos bit.

For the other bit, first you take 3 sin²- 3 cos² and turn it into - 3cos2Φ.
Then with the remaining 2 sin², you do the following
sin²=1-cos²
cos2Φ = 2cos²Φ - 1
:. ½(cos 2Φ + 1) = cos², giving us

(½ - ½cos2Φ ) + 3sin 2Φ - 3 cos2Φ.

a is therefore 3, b is -3.5 and c is 0.5

Can someone else check this this time please?
8. (Original post by Hitonagashi)
Crap!....I got c wrong I think... very embarrassing sorry.

c SHOULD read min -25...I somehow got the impression you were only between 0 and 90 with those limits...

For d i, think of the typical cos graph. It curves down, and then back up again between 0 and 360. If you draw a line anywhere going straight across the graph(like for example at 15 like in the question), it will cross the curve at 2 points, on the way down, and the reflection on the way up.

3sin2Φ is obviously the solution to the 6sincos bit.

For the other bit, first you take 3 sin²- 3 cos² and turn it into - 3cos2Φ.
Then with the remaining 2 sin², you do the following
sin²=1-cos²
cos2Φ = 2cos²Φ - 1
:. ½(cos 2Φ + 1) = cos², giving us

(½ - ½cos2Φ) + 3sin 2Φ - 3 cos2Φ.

a is therefore 3, b is -3.5 and c is 0.5

Can someone else check this this time please?
I wasted last 15minutes on this couldnt really get any where further, where did you get "3 sin²- 3 cos² " from?
9. 5sin²θ-3cos²θ+6sinθcosθ
2sin²θ+3sin²θ-3cos²θ +3sin2θ
2sin²θ-3(cos²θ- sin²θ ) + 3 sin 2θ
2sin²θ-3cos2Ø+3sin2θ

Actually nevermind I see where you got that from Hitonagashi, spitting the 5 into 3+2 its not a technique I've had to use before for a c3 trig question though.
10. Be prepared for it though...since it isnt new material, they could quite possibly ask that again on an exam paper...

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