Question

Trig identities, please help:

Please just teach me how to do a-d, But no answers:) I need to learn how to do them

Answer 1

trig ids for short

Answer 2

lol. yeah:)

Answer 3

@hero, are you there?

Answer 4

Yup

Answer 5

So is Mimi and Diya

Answer 6

first one:
\[\tan \theta = \frac{\sin \theta}{\cos \theta}\]
you'll get the answer

Answer 7

Yup:), first one i already had it. it was simple.

Answer 8

\[\sin^2 \theta + \cos^2 \theta = 1\]

Answer 9

(sinx+cosx)^2 expand using (a+b)^2 formula
and remember sin^x+cos^2x =1

Answer 10

@Mimi_x3 , your turn

Answer 11

Third..
\[\cos2x = 2\cos^{2} x -1\]

Answer 12

wait, so for the 2nd one, Im working on LHS

Answer 13

Also..
\[\sin^{2} x+\cos^{2} x = 1\]
for the third one..

Answer 14

\[(cosx+sinx)^2= \cos^2x+\sin^2x+2cosxsinx\]

Answer 15

@Mimi_x3 's post was like a hit and run

Answer 16

Want me to give the answe hero?

Answer 17

I didn't say that

Answer 18

So, (sinx+cosx)² becomes sin² x + 2sin*cosx+cos²x

Answer 19

right? for problem b

Answer 20

Put the sin^2x and the cos^2x together

Answer 21

Fourth one..
\[\tan^{2} x + 1 = \sec^{2} x \]
\[\tan^{2} x = \sec^{2} x - 1\]
then use trig identity of secx = 1/cosx
might workk..

Answer 22

from both ends right?

Answer 23

and then we get RHS

Answer 24

and sin^2x+cos^2x=1 so it'll be 1+2sinxcosx

2nd one^^ ok?

Answer 25

yes:)

Answer 26

oh, third one looks complicated

Answer 27

i gave you a hint try it; might work..

Answer 28

omg. im kind of lost, i dont even understand your hint. sorry:(

Answer 29

Ok wait

Answer 30

sin2x = 2sinxcosx
cotx=cosx/sinx

1+cos2x= cosx/sinx*2sinxcosx
=2cos^2x

Now?

Answer 31

Yes

Answer 32

ok:) yes. Thank you:)

Answer 33

2cos^2x=1+cos2x

Answer 34

okok. Im getting it abit:) that explanation helped:)

Answer 35

RHS: \[cotxsin2x \]
\[\frac{cosx}{sinx} *2sinxcosx = \frac{2sinxcos^{2x}}{sinx}
= \frac{2sinx(1-sin^{2}x)}{sinx}\]
\[2(1-\sin ^{2}x) = 2-2\sin ^{2}x\]
LHS:
\[1+\cos2x = \sin ^{2}x+\cos ^{2}x +(1-2\sin ^{2}x)\]
\[\sin ^{2}x +\cos ^{2}x + 1 - \sin ^{2}x\]
\[\sin ^{2}x +(1-\sin ^{2}x)+1-2\sin ^{2}x\]
\[2-2\sin ^{2}x \]

Answer 36

Therefore LHS = RHS

Answer 37

That really helps:) Thank you so much! Im sorry Im a bit slow... Im just not good at maths:(

Answer 38

But im trying hard to learn this

Answer 39

No problem O2S !!=)

Answer 40

Need help with the fourth one?

Answer 41

Yes please:)

Answer 42

Expanding the LHS might work..
\[\sin ^{2}x+\cos ^{2}x = 1 \]
\[\cos2x = 2\cos^{2} x -1\]

Answer 43

@Hero: Now it's your turn to help.

Answer 44

No Mimi, it's your turn until I say it's not :P