# More P2 triggles

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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

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

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#2

(Original post by

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

**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

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)

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(Original post by

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)

**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)

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#4

(Original post by

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...

**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

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#7

(Original post by

I know. I didn't

**It'sPhil...**)I know. I didn't

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?

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#8

(Original post by

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?

**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?

(x-1)/x

then multiply the 'x' to the numerator

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#9

**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?

Eg 1/(3x + 1/sinx) ... times everything by sinx to get sinx/(3x.sinx + 1).

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#10

**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

4/3+7=tan(A+B)(1-4/3.7)

tan(A+B)=-1

A+B=arctan(-1)

A+B=3pi/4.

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