# More P2 triggles

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#1
hey, I'm stuck on these questions, can anyone help?

firstly:
the acute angles A and B are such that tan A = 4/3 and tan B = 7
prove that A + B = 3pi / 4

I've drawn out triangles and labelled the sides, but I can't think of what else to do...

Secondly:
Prove that
2(sin^2(P)*sin^2(Q)) + 2(cos^2(P)*cos^2(Q)) = 1 + cos2P cos2Q

any suggestions?
cheers
0
15 years ago
#2
(Original post by kimoni)
hey, I'm stuck on these questions, can anyone help?

firstly:
the acute angles A and B are such that tan A = 4/3 and tan B = 7
prove that A + B = 3pi / 4

I've drawn out triangles and labelled the sides, but I can't think of what else to do...

Secondly:
Prove that
2(sin^2(P)*sin^2(Q)) + 2(cos^2(P)*cos^2(Q)) = 1 + cos2P cos2Q

any suggestions?
cheers
starting with rhs..
1+(cos^2P-sin^2P)(cos^2Q-sin^2Q))=1+cos^2Pcos^2Q+sin^2Psi n^2Q-sin^2Pcos^2Q-sin^2Qcos^2P

=1+cos^2Pcos^2Q+sin^2Psin^2Q+cos ^2Pcos^2Q-cos^2Q+sin^2Qsin^2Q-sin^2Q

=2(cos^2Pcos^2Q+sin^2Psin^2Q)+1-(cos^2Q+sin^2Q)
=2(cos^2Pcos^2Q+sin^2Psin^2Q)+1-1
=2(cos^2Pcos^2Q+sin^2Psin^2Q)
0
#3
(Original post by lgs98jonee)
starting with rhs..
1+(cos^2P-sin^2P)(cos^2Q-sin^2Q))=1+cos^2Pcos^2Q+sin^2Psi n^2Q-sin^2Pcos^2Q-sin^2Qcos^2P

=1+cos^2Pcos^2Q+sin^2Psin^2Q+cos ^2Pcos^2Q-cos^2Q+sin^2Qsin^2Q-sin^2Q

=2(cos^2Pcos^2Q+sin^2Psin^2Q)+1-(cos^2Q+sin^2Q)
=2(cos^2Pcos^2Q+sin^2Psin^2Q)+1-1
=2(cos^2Pcos^2Q+sin^2Psin^2Q)
Ah, cheers.. I didn't know you could start from the RHS, is it acceptable?
0
15 years ago
#4
(Original post by kimoni)
hey, I'm stuck on these questions, can anyone help?

firstly:
the acute angles A and B are such that tan A = 4/3 and tan B = 7
prove that A + B = 3pi / 4

I've drawn out triangles and labelled the sides, but I can't think of what else to do...

And its fine to start with RHS because the expression is tautologous -it is identical for all x
0
15 years ago
#5
(Original post by It'sPhil...)
tautologous
u cant make up words
0
15 years ago
#6
(Original post by lgs98jonee)
u cant make up words
I know. I didn't
0
15 years ago
#7
(Original post by It'sPhil...)
I know. I didn't
hehe

hey heres a question for ya.

when youve got a fraction in a fraction, how do you sort it out?

i mean things like 1-(1/x) on the bottom of the fraction... you can times the x up, but where else do you have to times it? the top too?
0
15 years ago
#8
(Original post by kikzen)
hehe

hey heres a question for ya.

when youve got a fraction in a fraction, how do you sort it out?

i mean things like 1-(1/x) on the bottom of the fraction... you can times the x up, but where else do you have to times it? the top too?
first make the 1-(1/x) a single fraction:
(x-1)/x
then multiply the 'x' to the numerator
0
15 years ago
#9
(Original post by kikzen)
hehe

hey heres a question for ya.

when youve got a fraction in a fraction, how do you sort it out?

i mean things like 1-(1/x) on the bottom of the fraction... you can times the x up, but where else do you have to times it? the top too?
What i do is times every single term by the fraction bit, including the top ones...

Eg 1/(3x + 1/sinx) ... times everything by sinx to get sinx/(3x.sinx + 1).
0
15 years ago
#10
(Original post by kimoni)
hey, I'm stuck on these questions, can anyone help?

firstly:
the acute angles A and B are such that tan A = 4/3 and tan B = 7
prove that A + B = 3pi / 4

I've drawn out triangles and labelled the sides, but I can't think of what else to do...

Secondly:
Prove that
2(sin^2(P)*sin^2(Q)) + 2(cos^2(P)*cos^2(Q)) = 1 + cos2P cos2Q

any suggestions?
cheers
tanA+tanB=tan(A+B)(1-tanAtanB)
4/3+7=tan(A+B)(1-4/3.7)
tan(A+B)=-1
A+B=arctan(-1)
A+B=3pi/4.
0
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