Maths extension 1 prelim question! (1 Viewer)

enigma_1 · Jul 13, 2013

Hey guys, can you please write solutions to these questions in the attached file?

PS Sorry I forgot to write the division signs, and that's why it looks like that.

Thank you so much!

HeroicPandas · Jul 13, 2013

Hyper_bole said:
Hey guys, can you please write solutions to these questions in the attached file?

PS Sorry I forgot to write the division signs, and that's why it looks like that.

Thank you so much!View attachment 28362

Trig identiy question:common denominate and then use do a reverse sine expansion (i.e. sinAcosB + sinBcosA = sin(A+B))

$\frac{sin3x}{sinx} + \frac{cos3x}{cosx}= \frac{sin3xcosx+cosx3xsinx}{sinxcosx} \\ \\ =\frac{sin4x}{\frac{1}{2}sin2x} \\ = \frac{2sin4x}{sin2x} \\ = \frac{2\cdot 2sin2xcosx}{sin2x} \\ = 4cos2x$
2ND TRIG Ideitntiy question:

$LHS = tan2xcotx \\ = \frac{2tanx}{1-tan^2x}\cdot \frac{1}{tanx} \\ = \frac{2}{1-(sec^2x-1)} \\ = \frac{2}{2-sec^2x} \\ = \frac{2cos^2x}{2cos^2x - 1} \\ = \frac{2cos^2x}{cos2x} \\ = \frac{1+cos2x}{cos2x} \\ = \sec(2x)+1$

LAST ONE: just use the sine and cosine expansion, as well as double angles

$sin3x = sin(2x+x) \\ = sin2xcosx + cos2xsinx \\ = 2sinxcos^2x+(cos^2x-sin^2x)sinx \\ = 2sinx(1-sin^2x)+ cos^2xsinx - sin^3x \\ =2sinx-2sin^3x + sinx(1-sin^2x) - sin^3x \\ = 3sinx - 4sin^3x$

$cos3x = cos(2x+x) \\ = cos2xcosx - sin2xsinx \\ = (2cos^2x-1)cosx - 2(1-cos^2x)cosx\\ = 4cos^3x - 3cosx$

Maths extension 1 prelim question! (1 Viewer)

enigma_1

Miss Cricket

HeroicPandas

Heroic!

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Maths extension 1 prelim question! (1 Viewer)

enigma_1

~~~~ Miss Cricket ~~~~

HeroicPandas

Heroic!

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Miss Cricket