The Student Room Group

A fairly basic question

I know this is something that is probably quite obvious to most of you, but it always confuses me.

We know this is true: sin2xcos2x=tan2x \frac{sin^2 x}{cos^2 x} =tan^2 x

But using the laws of indices, we know that dividing one exponent by another is the same as subtracting one of the exponents from the other. So why are we left with a trigonometric function raised to the same power?
Reply 1
That's only true with the same base. sin is not cos so you can't change the index. So 2^9 ÷ 2^4 = 2^5, but you can't say that 6^9 ÷ 2^4 is 6^5.
(edited 5 years ago)
Reply 2
Original post by Sinnoh
That's only true with the same base. sin is not cos.


True. So it is just something i'm just supposed to know, rather than understand?
Reply 3
Becuase sin^2(x)/cos^2(x) is equal to (sin(x)/cos(x))^2

Sin(x)/Cos(x) = tan(x) .......... so (sin(x)/cos(x))^2 = (tan(x))^2 = tan^2(x)
Original post by Illidan2
True. So it is just something i'm just supposed to know, rather than understand?


Well this is something you understand from like Y9 GCSE so nobody bothers explaining at this level because it's so obvious.

We have

Unparseable latex formula:

\begin{aligned} \dfrac{x^n}{y^n} & = \dfrac{ \overbrace{x \cdot x \cdot x \ldots \cdot x}^{\text{n times}}} {\underbrace{y \cdot y \cdot y \ldots \cdot y}_{\text{n times}}} \\ & \\ & = \underbrace{\frac{x}{y} \cdot \frac{x}{y} \cdot \ldots \cdot \frac{x}{y}}_{\text{n times}} \\ & \\ & =\left( \frac{x}{y} \right)^n



Letting x=sinθx=\sin \theta and y=cosθy=\cos \theta with n=2n=2 achieves what you're confused about.
(edited 5 years ago)
Reply 5
Original post by Illidan2
True. So it is just something i'm just supposed to know, rather than understand?


(2*2*2*2*2)÷(2*2*2) = 2*2. That's understandable.
Reply 6
Original post by Sinnoh
(2*2*2*2*2)÷(2*2*2) = 2*2. That's understandable.


No, I know. I understood you the first time, and I understood RDKGames straight away when he explained it. I simply didn't make the connection that I could treat trig functions the same way, when, of course, I can. It's just that I don't always make a connection between the patterns e.g. I know that (x/y)^2= (x^2/y^2), but I didn't make that connection when it was in trig function format, with sin and cos as x and y.

Quick Reply

Latest