Trig pls help (2 Viewers)

[wandering17]

How do i prove this identity?

cos(A-B)-cos(A+B) / Sin(A+B) +Sin(A-B) = TanB

 

[Squar3root]

expanding everything and clean up

 

[wandering17]

show me please?

 

[Rhinoz8142]

show me please?

just try it by looking at ur notes and textbook then show us ur working out.. then we will see.



Nothing comes in a silver platter m8

 

[hypermax]

show me please?

I hope you dont take this to offence but are you familiar with compound angles ( the identities ).

 

[wandering17]

yeh

cosAcosB + SinAsinB - CosAcosB - SinAsinB
-------------------------------------------------
SinAcosB+CosAsinB + SinAcosB-CosAsinB

Top line, don't they just all cancel out? then im left with the bottom of 2sinAcosB
then how do i get to tanB

 

[Librah]

yeh

cosAcosB + SinAsinB - (CosAcosB - SinAsinB)
-------------------------------------------------
SinAcosB+CosAsinB + SinAcosB-CosAsinB

Top line, don't they just all cancel out? then im left with the bottom of 2sinAcosB
then how do i get to tanB

Watch your order of operations on top line.

 

[FrankXie]

yeh

cosAcosB + SinAsinB - CosAcosB - SinAsinB
-------------------------------------------------
SinAcosB+CosAsinB + SinAcosB-CosAsinB

Top line, don't they just all cancel out? then im left with the bottom of 2sinAcosB
then how do i get to tanB

top line, last term, should be plus sign instead of minus

 

[wandering17]

top line, last term, should be plus sign instead of minus

oooohhh. Thanks, my bad

 

[faisalabdul16]

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{cosAcosB&space;&plus;&space;sinAsinB&space;-&space;(cosAcosB-sinAsinB)}{sinAcosB&space;&plus;cosAsinB&space;&plus;&space;sinAcosB-&space;cosAsinB}&space;\\&space;\\&space;\\&space;=&space;\frac{2sinAsinB}{2sinAcosB}&space;\\&space;\\&space;\\&space;=tanB" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{cosAcosB&space;&plus;&space;sinAsinB&space;-&space;(cosAcosB-sinAsinB)}{sinAcosB&space;&plus;cosAsinB&space;&plus;&space;sinAcosB-&space;cosAsinB}&space;\\&space;\\&space;\\&space;=&space;\frac{2sinAsinB}{2sinAcosB}&space;\\&space;\\&space;\\&space;=tanB" title="\frac{cosAcosB + sinAsinB - (cosAcosB-sinAsinB)}{sinAcosB +cosAsinB + sinAcosB- cosAsinB} \\ \\ \\ = \frac{2sinAsinB}{2sinAcosB} \\ \\ \\ =tanB" /></a>

 

[Speed6]

yeh

cosAcosB + SinAsinB - CosAcosB - SinAsinB
-------------------------------------------------
SinAcosB+CosAsinB + SinAcosB-CosAsinB

Top line, don't they just all cancel out? then im left with the bottom of 2sinAcosB
then how do i get to tanB

BE CAREFUL (order of operations) + --> - (+ on top)---(- on bottom)

 

[Kaido]

gg in the exam. we've all done it b4 trollface.jpg

 

[wandering17]

I have another question... if anyone could please help me, idk how to approach the question at all?

Use the identity cos2theta = 1-2sin^2theta to find sin^2 15

How do i start this?

 

[Kaido]

well, rearrange to isolate sin^2feta and sub in fatter=15

 

[FrankXie]

I have another question... if anyone could please help me, idk how to approach the question at all?

Use the identity cos2theta = 1-2sin^2theta to find sin^2 15

How do i start this?

in that identity, put theta=15 degree