1. ## questions about trigo and circles

don't even know how to begin, does it involve factor/sum formulae?

angle ABC and angle ADC is 90 deg respectively. because AC is diameter.
how do i relate them and find AD?

sin x cos x = -(3/10)
then ....can i do this :

sin x = - (3/10) , cos x = - (3/10)

??? or i can only do this when it's equal to zero?

what i tried to do:

centre(0,0)
thus, i roughly know it would be something like (x^2)+(y^2).....
A(a,0)
B(0,b)
AB=2sqrt2

so..

sqrt{[(a-0)^2]+[(0-b)^2]}=2sqrt2
(a^2)+(b^2)=8

and then......i'm stuck ="=
how am i going to find the radius ...

2. 1. Use the identities:
$\sin{A}-\sin{B} = 2\cos{\left(\frac{A+B}{2}\right)}\sin{\left(\frac{ A-B}{2}\right)}$
$\cos{A}+\cos{B}=2\cos{\left(\frac{A+B}{2}\right)}\ cos{\left(\frac{A-B}{2}\right)}$

to verify that [1]/[2] is indeed the identity in question.

===========

[1] $\sin{x}-\sin{y} = 2\cos{\left(\frac{x+y}{2}\right)}\sin{\left(\frac{ x-y}{2}\right)}$
[2] $\cos{x} + \cos{y} = 2\cos{\left(\frac{x+y}{2}\right)}\cos{\left(\frac{ x-y}{2}\right)}
$

$\implies \frac{2\cos{\left(\frac{x+y}{2}\right)}\sin{\left( \frac{x-y}{2}\right)}}{2\cos{\left(\frac{x+y}{2}\right)}\c os{\left(\frac{x-y}{2}\right)}} = \frac{\sin{\left(\frac{x-y}{2}\right)}}{\cos{\left(\frac{x-y}{2}\right)}} = \tan{\left(\frac{x-y}{2}\right)}$

Surely you know what to do next?

3. 1) $\sin x-\sin y=2\sin\frac{x-y}{2}\cos\frac{x+y}{2}$
$\cos x+\cos y=2\cos\frac{x-y}{2}\cos\frac{x+y}{2}$

2) Let $\widehat{BCA}=\alpha, \ \widehat{ACD}=\beta$
$AC=\sqrt{AB^2+BC^2}=\sqrt{85}$
$\sin\alpha=\frac{AB}{AC}=\frac{7}{\sqrt{85}}, \ \cos\alpha=\frac{BC}{AC}=\frac{6}{\sqrt{85}}$

Now, $\sin\beta=\sin(\theta-\alpha)=\sin\theta\cos\alpha-\sin\alpha\cos\theta=\frac{6}{\sqrt{85}}\sin\theta-\frac{7}{\sqrt{85}}\cos\theta$
But $\sin\beta=\frac{AD}{AC}=\frac{AD}{\sqrt{85}}$

4. 3) $10\sin x\cos x=5\cdot 2\sin x\cos x=5\sin 2x$

5. $A(a,0), \ B(0,a)$
Try again and find a.

6. Hello, wintersoltice!

Given: . $\begin{array}{cccc}{\color{blue}[1]} &\sin x - \sin y &=&\frac{1}{2} \\ \\[-4mm] {\color{blue}[2]} & \cos x + \cos y &=& \frac{1}{4} \end{array}$ . find the value of: . $\tan\left(\tfrac{x-y}{2}\right)$

. . $\text{(i) }\;\tfrac{1}{8}\qquad\text{(ii) }\;0 \qquad\text{(iii) }\;\tfrac{1}{2}\qquad\text{(iv) }\;2$
We need two Sum-to-Product identities:

. . $\sin A - \sin B \:=\:2\cos\left(\tfrac{A+B}{2}\right)\sin\left(\tf rac{A-B}{2}\right)$

. . $\cos A + \cos B \:=\:2\cos\left(\tfrac{A+B}{2}\right)\cos\left(\tf rac{A-B}{2}\right)$

Then [1] and [2] become: . $\begin{array}{ccccc}
2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) &=& \frac{1}{2} & {\color{blue}[3]} \\ \\[-3mm]
2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) &=& \frac{1}{4} & {\color{blue}[4]} \end{array}$

Divide [3] by [4]: . $\frac{2\cos\left(\frac{x+y}{2}\right)\sin\left(\fr ac{x-y}{2}\right)}{2\cos\left(\frac{x+y}{2}\right)\cos\ left(\frac{x-y}{2}\right)} \;=\;\frac{\frac{1}{2}}{\frac{1}{4}}$ . $\Rightarrow\quad \frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} \;=\;2$

. . Therefore: . $\tan\left(\frac{x-y}{2}\right) \:=\:2$ . . . answer (iv)

Edit: Too slow again . . .
.

7. Originally Posted by red_dog
3) $10\sin x\cos x=5\cdot 2\sin x\cos x=5\sin 2x$

10 sin x cos x +3 =0
(5)(2)sin x cos x =-3
5 sin 2x =-3
sin 2x = -(3/5)
basic angle of 2x = 36.87 deg
90deg < x < 180 deg .......=> second quadrant
therefore,
2x= (180deg - 36.87deg)
2x= 143.13 deg
x= 71.565 deg

>"<....
there's no this answer among the choices.....
where did i went wrong ?? @"@
based on my sixth sense, i feel that the part i higlighted in bold is incorrect/weird/etc(don't know how to describe)..........

Originally Posted by red_dog
$A(a,0), \ B(0,a)$
Try again and find a.
why you know they are the same???
i mean the unknown, a.

8. Originally Posted by wintersoltice
1

why you know they are the same???
i mean the unknown, a.
Because OA and OB are radius in the circle, so OA=OB.