help... maths (1 Viewer)

B1andB2 · Jun 17, 2020

Hi, can someone please help me with this q on compound angles

If secx-tanx=3/5 , show that sinx=8/17 when x is acute

Thankk you

Drdusk · Jun 17, 2020

B1andB2 said:
Hi, can someone please help me with this q on compound angles

If secx-tanx=3/5 , show that sinx=8/17 when x is acute

Thankk you

$\text{So you have} \hspace{2mm}\dfrac{1}{\cos x} - \dfrac{\sin x}{\cos x} = \dfrac{3}{5}$

$\dfrac{1- \sin x}{\cos x} = \dfrac{3}{5}$

$\text{Square both sides and get everything in terms of sin x using the identity}$
$\sin ^2 x + \cos ^2 x = 1$

$\text{It's straight-forward from there. }$

B1andB2 · Jun 17, 2020

Drdusk said:
$\text{So you have} \hspace{2mm}\dfrac{1}{\cos x} - \dfrac{\sin x}{\cos x} = \dfrac{3}{5}$

$\dfrac{1- \sin x}{\cos x} = \dfrac{3}{5}$

$\text{Square both sides and get everything in terms of sin x using the identity}$
$\sin ^2 x + \cos ^2 x = 1$

$\text{It's straight-forward from there. }$

omg thank you so muchhh! i didn't think of the squaring nice!

CM_Tutor · Jun 17, 2020

B1andB2 said:
Hi, can someone please help me with this q on compound angles

If secx-tanx=3/5 , show that sinx=8/17 when x is acute

Thankk you

Take as equation (1): $\sec x - \tan x = \frac{3}{5}$

Multiplying (1) by $\sec x + \tan x$ gives equation (2):
$\sec^2 x - \tan^2 x = \frac{3}{5} (\sec x + \tan x) \implies \sec x + \tan x = \frac{5}{3}$

(1) + (2) gives $2\sec x = \frac{34}{15} \implies \cos x = \frac{15}{17}$

(2) - (1) gives $2\tan x = \frac{16}{15} \implies \tan x = \frac{8}{15}$

$\tan x = \frac{\sin x}{\cos x} \implies \sin x = \cos x \tan x = \frac{15}{17} \times \frac{8}{15} = \frac{8}{17}$

B1andB2 · Jun 17, 2020

CM_Tutor said:
Take $\sec x - \tan x = \frac{3}{5}$ as equation (1).

Multiplying (1) by $\sec x + \tan x$ gives that $\sec x + \tan x = \frac{5}{3}$ , call that equation (2).

(1) + (2) gives $2\sec x = \frac{34}{15} \implies \cos x = \frac{15}{17}$

(2) - (1) gives $2\tan x = \frac{16}{15} \implies \tan x = \frac{8}{15}$

$\tan x = \frac{\sin x}{\cos x} \implies \sin x = \cos x \tan x = \frac{15}{17} \times \frac{8}{15} = \frac{8}{17}$

oh wow, that's a cool method! Thanks!

CM_Tutor · Jun 17, 2020

Drdusk said:
$\text{So you have} \hspace{2mm}\dfrac{1}{\cos x} - \dfrac{\sin x}{\cos x} = \dfrac{3}{5}$

$\dfrac{1- \sin x}{\cos x} = \dfrac{3}{5}$

$\text{Square both sides and get everything in terms of sin x using the identity}$
$\sin ^2 x + \cos ^2 x = 1$

$\text{It's straight-forward from there. }$

You can also rewrite

$\dfrac{1 - \sin x}{\cos x} = \dfrac{3}{5}$

as

$1 - \sin x = \dfrac{3}{5} \cos x \qquad \text{(1)}$

Then, by noting that

$\dfrac{1 - \sin x}{\cos x} = \dfrac{\cos x}{1 + \sin x}$

and thus that RHS is also 3/5, you can show that

$1 + \sin x = \dfrac{5}{3} \cos x \qquad \text{(2)}$

Adding (1) and (2) gives $\cos x$ which can be back-substituted into either equation to get the desired result.

help... maths (1 Viewer)

B1andB2

oui oui baguette

Drdusk

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B1andB2

oui oui baguette

CM_Tutor

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B1andB2

oui oui baguette

CM_Tutor

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