# Questions (1 Viewer)

#### 181jsmith

##### Member
1. Solve the equation $\bg_white \sqrt{2}cos(x+60)-1=0 for 0\leq x\leq 360$

2. P (2ap, ap^2) and Q (2aq, aq^2) are 2 points on the parabola x^2= 4ay with parameter values p=4 and q=-6. Show that the lines OP and OQ are inclined at 45 degrees to each other.

3. Solve the inequality $\bg_white \frac{1}{x} - \frac{1}{x-2}> 0$

4. Show that 2cos(A - B)sin(A+B)= sin2A + sin 2B

5. The polynomial eq $\bg_white x^{3} + bx^{2} + cx +d =0$ has roots
a, a^2 and a^3

Find in terms of b,c and d $\bg_white \frac{1}{a}+\frac{1}{a^{2}}+\frac{1}{a^{3}}$

Show that $\bg_white b^{3}d-c^{3}=0$

thanks.

#### SpiralFlex

##### Well-Known Member
1. $\bg_white \cos (x+60)=\frac{1}{\sqrt{2}}$

$\bg_white x+60 = 315, 390$

$\bg_white x=255, 330$

2. $\bg_white P(8a, 16a), Q(-12,36a), O(0,0)$

Find the gradients, you should get $\bg_white 2, -3$ respectively.

$\bg_white PO: y=2x, QO: y=-3x$

$\bg_white \tan \theta = |\frac{2-(-3)}{1+3*-2}|$

$\bg_white \therefore \theta = 45$

3. $\bg_white \frac{x-2-x}{x(x-2)}>0$

$\bg_white \frac{-2}{x(x-2)}>0$

$\bg_white -2x(x-2)>0$

$\bg_white 2

4. $\bg_white =(\cos A \cos B + \sin A \sin B)(\sin A \cos B + \cos A \sin B)$

$\bg_white =\cos A \cos^2 B \sin A +\cos^2 A \cos B \sin B +\sin^2 A \sin B \cos B + \sin A \sin^2 B \cos A$

$\bg_white =(\sin^2 A + \cos^2 A)(2 \sin B \cos B)+(\sin^2 B + \cos^2 B)(2\sin A \cos A)$

$\bg_white =\sin 2A + \sin 2B$

5 i. Sum of roots: $\bg_white a+a^2+a^3=-b...(1)$

Two at a time: $\bg_white a^3+a^4+a^5=c...(2)$

Product: $\bg_white a^6=-d...(3)$

$\bg_white \frac{(2)}{(1)}$

$\bg_white a^2=-\frac{c}{b}$

$\bg_white \frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}=\frac{a^2(a+a^2+a^3)}{a^6}$

$\bg_white =\frac{\sqrt[3]{-d}*-b}{(-\frac{c}{b})^3}$

$\bg_white =\frac{\sqrt[3]{-d}*b^4}{c^3}$

ii. $\bg_white (a^2)^3=a^6$

$\bg_white (-\frac{c}{b})^3=-d$

$\bg_white b^3d-c^3=0$

#### taeyang

##### Member
1. $\bg_white \cos (x+60)=\frac{1}{\sqrt{2}}$

$\bg_white x+60 = 315, 390$

$\bg_white x=255, 330$

2. $\bg_white P(8a, 16a), Q(-12,36a), O(0,0)$

Find the gradients, you should get $\bg_white 2, -3$ respectively.

$\bg_white PO: y=2x, QO: y=-3x$

$\bg_white \tan \theta = |\frac{2-(-3)}{1+3*-2}|$

$\bg_white \therefore \theta = 45$

3. $\bg_white \frac{x-2-x}{x(x-2)}>0$

$\bg_white \frac{-2}{x(x-2)}>0$

$\bg_white -2x(x-2)>0$

$\bg_white 2

4. $\bg_white =(\cos A \cos B + \sin A \sin B)(\sin A \cos B + \cos A \sin B)$

$\bg_white =\cos A \cos^2 B \sin A +\cos^2 A \cos B \sin B +\sin^2 A \sin B \cos B + \sin A \sin^2 B \cos A$

$\bg_white =(\sin^2 A + \cos^2 A)(2 \sin B \cos B)+(\sin^2 B + \cos^2 B)(2\sin A \cos A)$

$\bg_white =\sin 2A + \sin 2B$

5 i. Sum of roots: $\bg_white a+a^2+a^3=-b...(1)$

Two at a time: $\bg_white a^3+a^4+a^5=c...(2)$

Product: $\bg_white a^6=-d...(3)$

$\bg_white \frac{(2)}{(1)}$

$\bg_white a^2=-\frac{c}{b}$

$\bg_white \frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}=\frac{a^2(a+a^2+a^3)}{a^6}$

$\bg_white =\frac{\sqrt[3]{-d}*-b}{(-\frac{c}{b})^3}$

$\bg_white =\frac{\sqrt[3]{-d}*b^4}{c^3}$

ii. $\bg_white (a^2)^3=a^6$

$\bg_white (-\frac{c}{b})^3=-d$

$\bg_white b^3d-c^3=0$
Juices are flowing.

#### bleakarcher

##### Active Member
for q5 i) can u just write -c/d as an equivalent expression or does it have to be in terms of b also

#### SpiralFlex

##### Well-Known Member
I would assume by the question, in terms of "b, c, d".