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# oh, easy trig watch

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1. angle B is acute and tanB = 4/3.

Find sinB
2. Use SOH CAH TOA

If tan B is 4/3, that means that the side of the triangle opposite to the angle is of relative length 4 and the side of the triangle adjacent to the angle is of relative length 3.

This means that by Pythagoras' the hypotenuse is root of 4^2 + 3^2, which is 5.

Hence, sin B is opposite over hypotenuse, which is 4/5.
3. umm you can just use a 3,4,5 triangle...
so if you draw that...
sinB = 4/5 (opp/hyp)
so just inverse sin(4/5) and should give a number?

(someone kick me if im doing it completely wrong )
4. omg... ignore me.
5. cant we just kick you instead?
6. ok then, let's try this...

angle B is acute and tanB = 3/4

Find sin(B/2)
7. (Original post by chewwy)
angle B is acute and tanB = 4/3.

Find sinB

same route as the others, expressed slightly differently:
tanb = 4/3
sinb/cosb=4/3
3sinb=4cosb
9sin²b=16cos²b
9sin²b=16(1-sin²b)
9sin²b=16-16sin²b
25sin²b=16
sin²b=16/25
sinb=4/5

Aitch
8. (Original post by Aitch)
same route as the others, expressed slightly differently:
tanb = 4/3
sinb/cosb=4/3
3sinb=4cosb
9sin²b=16cos²b
9sin²b=16(1-sin²b)
9sin²b=16-16sin²b
25sin²b=16
sin²b=16/25
sinb=4/5

Aitch
There are two flaws in this argument that cancel each other out (who said that two wrongs make a right?). I've emboldened them.

First one: Squaring will give extra solutions. You should check all solutions in the original equation.

Second one: Rooting will give positive and negative values (i.e. ±4/5)
9. (Original post by chewwy)
ok then, let's try this...

angle B is acute and tanB = 3/4

Find sin(B/2)
^^^
10. cos(A+B)=cosAcosB-sinAsinB

Let A and B both=B/2, therefore cosB=cos^2(B/2)-sin^2(B/2).

Everybody's favourite identity says cos^2(B/2)=1-sin^2(B/2).

This means cosB=1-2sin^2(B/2). From the above posts, we can see that cosB is 3/5.

So 3/5=1-2sin^2(B/2), sin^2(B/2)=1/5, sin(B/2)=+ or - 1/root5.

Since it's acute, choose the + value.
11. (Original post by chewwy)
ok then, let's try this...

angle B is acute and tanB = 3/4

Find sin(B/2)
sin²(B/2) = ½(1 - cosB)
sin²(B/2) = ½(1 - 3/5)
sin²(B/2) = ½(2/5)
sin²(B/2) = 1/5
sin(B/2) = 1/√5
===========
12. Aaaarg! just beat
13. (Original post by JohnSPals)
There are two flaws in this argument that cancel each other out (who said that two wrongs make a right?). I've emboldened them.

First one: Squaring will give extra solutions. You should check all solutions in the original equation.

Second one: Rooting will give positive and negative values (i.e. ±4/5)
B is acute?

Aitch
14. (Original post by Fermat)
sin²(B/2) = ½(1 - cosB)
sin²(B/2) = ½(1 - 3/5)
sin²(B/2) = ½(2/5)
sin²(B/2) = 1/5
sin(B/2) = 1/√5
===========
thankyou. i don't mean to be picky, but neither of you noticed that while tanB was 4/3 in my first question, it was 3/4 in the second, hence cosB was 4/5, and the answer is sqrt10/10
15. (Original post by chewwy)
thankyou. i don't mean to be picky, but neither of you noticed that while tanB was 4/3 in my first question, it was 3/4 in the second, hence cosB was 4/5, and the answer is sqrt10/10
good stuff, but a bit tricky at this time of night after 8 glasses of claret...

Aitch
16. (Original post by chewwy)
thankyou. i don't mean to be picky, but neither of you noticed that while tanB was 4/3 in my first question, it was 3/4 in the second, hence cosB was 4/5, and the answer is sqrt10/10
We were just making sure you were checking the answers and not copying them down straight off TSR into your homework...

I should be a teacher
17. (Original post by Aitch)
B is acute?

Aitch
Ahh I missed that .
18. (Original post by sarforaz)
cant we just kick you instead?
nuff of the high school banter...
Do you guys want a ****ing blue ticket? ^o)
(visesh)
19. (Original post by Camford)
nuff of the high school banter...
Do you guys want a ****ing blue ticket? ^o)
(visesh)
ah man.

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