Trigonometry: count sin+cos when tg-ctg=-7/12

[MrGoATi]

Mentor note: Moved thread to homework section

ok So I'm doing supposedly easy trigonometry problems. i did the easiest ones. now I have no idea how to solve 2.

first one is count
sin+cos
When
tg - (1/tg) = -(7/12)
what i figured is that i probably need to use (sin/cos) - (cos/sin) = -7/12
but i came empty after 30 minutes of playing with both formulas.

The ASNWER is supposed to be 1.4
************
the other problem is prove:
cosx^2=siny*cosy
when
cos(π+2x)=2sin^2(π/4-y)

the second formula I made into
cos(2x)=2cos^2(y)
then used cos(2x) = 2cos^2(x) -1
cosx^2 = ( 2cos^2(y)+1 ) / 2

if all that's correct. Then it made it possible to connect both formulas leaving just y.
( 2cos^2(y)+1 ) / 2 = siny*cosy
but i can only go in circles here

 

[BvU]

(sin/cos) - (cos/sin) = -7/12

How would you solve $x-{1\over x}={-7\over 12}$ ?

 

[MrGoATi]

How would you solve $x-{1\over x}={-7\over 12}$ ?

Forgot to say the radiant is between [0;90]
oh, yeah. x² + 7x/12 - 1 = 0
got
x₁=4/3
x₂= - 3/4

my guess since it's tg and [0;90] is positive. tg = 4/3 and not -3/4 right?
i know i can do sin = tgcos;
so tgcos+cos
=>(7/3)cos

used tg² = 1/cos²
cos = 3/5

and got the correct answer (7/3)*(3/5) = 7/5= 1.4
Thank you very much! :)
**********************

i will probably have to make similar problems with limited time, and without having correct answer so can you tell me if my way of choosing between
x₁=4/3
x₂= - 3/4
was correct, or should i try with both or what?

******
and would you or someone else help with second problem?

 

[Dick]

and would you or someone else help with second problem?

On the second problem, you've got a problem from the start. The first simplification isn't right. $\cos(\pi+2x)=2\sin^2(\pi/4-y)$ does not imply $\cos(2x)=\cos^2(y)$. Try that again.

 

[MrGoATi]

On the second problem, you've got a problem from the start. The first simplification isn't right. $\cos(\pi+2x)=2\sin^2(\pi/4-y)$ does not imply $\cos(2x)=\cos^2(y)$. Try that again.

it's cos2x= -sin²(y) right?

It does not help much though.

 

[Dick]

it's cos2x= -sin²(y) right?

It does not help much though.

No, that's not right either. If you are not just guessing then I suggest you show your steps. Use the trig sum identities.

 

[MrGoATi]

No, that's not right either. If you are not just guessing then I suggest you show your steps. Use the trig sum identities.

yeah it kinda worked, I came real close but one additional - is there and i can't find a mistake..
so with first equation i make it
sinycosy=(1+cos2x)/2

with the second.
cos2x=2(sin(π/4)cosy-sinycos(π/4))²
both sin(π/4) and sin(π/4)=root2/2
and when squared = ½
cos2x=2(½sin²+½cos²-sincos)
cos2x=1-2sincos
-(cos2x-1)/2=sinycosy

so it is very similar but not same
sinycosy=-(cos2x-1)/2
sinycosy=(cos2x+1)/2

any idea what I'm still doing wrong?

 

[Dick]

yeah it kinda worked, I came real close but one additional - is there and i can't find a mistake..
so with first equation i make it
sinycosy=(1+cos2x)/2

with the second.
cos2x=2(sin(π/4)cosy-sinycos(π/4))²
both sin(π/4) and sin(π/4)=root2/2
and when squared = ½
cos2x=2(½sin²+½cos²-sincos)
cos2x=1-2sincos
-(cos2x-1)/2=sinycosy

so it is very similar but not same
sinycosy=-(cos2x-1)/2
sinycosy=(cos2x+1)/2

any idea what I'm still doing wrong?

I'm guessing that you seem to think that $\cos(\pi+2x)=\cos(2x)$. That's not right. Use the sum rules again.