Trig help (1 Viewer)

dabatman · May 6, 2017

Hi can anyone help with these questions from the 3 unit cambridge book

Question d requires to eliminate theta and the rest involves proving identities.

Thank you

He-Mann · May 6, 2017

What is sin²x + cos²x equal to?

Mahan1 · May 6, 2017

dabatman said:
Hi can anyone help with these questions from the 3 unit cambridge book

Question d requires to eliminate theta and the rest involves proving identities.

Thank you

$\textrm{Note in part d} \ x^{2} = sin^{2}\theta + 2sin\theta \ cos\theta + cos^{2}\theta$

$\textrm{and} \ y^{2} = sin^{2}\theta -2sin\theta \ cos\theta + cos^{2}\theta$

$\textrm{Hence}\ x^{2} + y^{2} = 2(sin^{2}\theta + cos^{2}\theta) = 2$

$\textrm{Second method is to observe} \ x+y = 2 sin\theta, x-y = 2cos\theta \ \textrm{then square them, see what happens.}$

$\textrm{Third method is to use the following identities:} \ cos^{2}\theta - sin^{2}\theta = cos2\theta = 2cos^{2}\theta -1$

$\textrm{Note} -xy = cos^{2}\theta -sin^{2}\theta = cos2\theta = 2cos^{2}\theta -1 = \frac{(x-y)^{2}}{2} -1$

$\textrm{then you get} \ -xy = \frac{(x-y)^{2}}{2} -1 \ \textrm{then simplify it.}$

$\textrm{For part e, you need to use the following identity} \ 1+tan^{2}x = sec^{2}x$

$\textrm{So} \ LHS = \frac{1}{1+tan^{2}} - \frac{1}{1+sec^{2}x} = \frac{1}{sec^{2}} - \frac{1}{1+sec^{2}x}$

$= cos^{2}x - \frac{cos^{2}x}{cos^{2}x+1} = cos^{2}x(1-\frac{1}{1+cos^{2}x}) = cos^{2}x(\frac{cos^{2}x}{1+cos^{2}x})$

$= \frac{cos^{4}x}{1+cos^{2}x} = RHS$

$\textrm{Second method, just write tan and sec in terms of sine and cosine}$

$tanx = \frac{sinx}{cosx}, secx = \frac{1}{cosx} \ \textrm{I leave it to you to check}$

$\textrm{for part f,} \ LHS = \frac{cos\theta}{1-tan\theta} + \frac{sin\theta}{1-cot\theta} = \frac{cos^{2}\theta}{cos\theta - sin\theta} + \frac{sin^{2}\theta}{sin\theta - cos\theta}$

$= \frac{cos^{2}x - sin^{2}x}{cosy -sinx} = \frac{(cosx - sinx)(cosy + sinx)}{cosx -sinx} = cosy + sinx$

boredofstudiesuser1 · May 7, 2017

He-Mann said:
What is sin²x + cos²x equal to?

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