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Trigonometric identities?

I need to simplify

tan(x)*√(1-sin⁻²(x))

the only two identities I can use are tan(x)=sin(x)/cos(x) and sin²(x)+cos²(x)=1 (these are the only identities in the specification of the course I’m studying)

How would I get rid of the square root?
Original post by annabel_gilmour
I need to simplify

tan(x)*√(1-sin⁻²(x))

the only two identities I can use are tan(x)=sin(x)/cos(x) and sin²(x)+cos²(x)=1 (these are the only identities in the specification of the course I’m studying)

How would I get rid of the square root?

From the identity sin^2 + cos^2 = 1, you can notice that you will get sin⁡−2x\sin^{-2}x if you proceed to divide both sides of it by sin⁡2x\sin^2 x.
Original post by annabel_gilmour
I need to simplify

tan(x)*√(1-sin⁻²(x))

the only two identities I can use are tan(x)=sin(x)/cos(x) and sin²(x)+cos²(x)=1 (these are the only identities in the specification of the course I’m studying)

How would I get rid of the square root?


Is that expression written correctly, because 1−1sin⁡2(x)1-\frac{1}{\sin^2(x)} is negative for real xx?
Reply 3
Avatar for S.78_t
S.78_t
12
Original post by annabel_gilmour
I need to simplify

tan(x)*√(1-sin⁻²(x))

the only two identities I can use are tan(x)=sin(x)/cos(x) and sin²(x)+cos²(x)=1 (these are the only identities in the specification of the course I’m studying)

How would I get rid of the square root?


The square root is equivalent to the power of 1/2
So it is tan(x)*(1-sin^-2 (x))^1/2
Original post by 15Characters...
Is that expression written correctly, because 1−1sin⁡2(x)1-\frac{1}{\sin^2(x)} is negative for real xx?

I don’t know :frown: I’ve typed up exactly what the textbook says but it’s a new textbook and I have found some mistakes as I’ve been going along. The answer in the back is sinx...
Original post by RDKGames
From the identity sin^2 + cos^2 = 1, you can notice that you will get sin⁡−2x\sin^{-2}x if you proceed to divide both sides of it by sin⁡2x\sin^2 x.

I’m probably being thick but can you clarify what you mean by both sides? It’s an expression so there’s only one side I think
Reply 6
Avatar for davros
davros
16
Original post by annabel_gilmour
I don’t know :frown: I’ve typed up exactly what the textbook says but it’s a new textbook and I have found some mistakes as I’ve been going along. The answer in the back is sinx...

Are you sure the question isn't tan⁡x⋅1−sin⁡2x\tan x \cdot \sqrt{1 - \sin^2 x} ?
Original post by davros
Are you sure the question isn't tan⁡x⋅1−sin⁡2x\tan x \cdot \sqrt{1 - \sin^2 x} ?

BE717BE4-9236-4EA4-A557-FFD9ABAB0A99.jpeg
this is a picture of the textbook - it could be a mistake though?
Reply 8
Avatar for davros
davros
16
Original post by annabel_gilmour
BE717BE4-9236-4EA4-A557-FFD9ABAB0A99.jpeg
this is a picture of the textbook - it could be a mistake though?

Well if you write tan x = sin x / cos x and then take the sin x inside the square root (so it becomes a multiplier of sin^2 x you can see that something strange is going on :smile:
in

Original post by davros
Well if you write tan x = sin x / cos x and then take the sin x inside the square root (so it becomes a multiplier of sin^2 x you can see that something strange is going on :smile:

Shall I just leave this question then?
Reply 10
Avatar for davros
davros
16
Original post by annabel_gilmour
in


Shall I just leave this question then?

To me it just looks like a misprint so yes :smile:

Is anyone else in your class trying the same exercise?
Original post by davros
To me it just looks like a misprint so yes :smile:

Is anyone else in your class trying the same exercise?

No - I’m self teaching so it’s just me!

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