x² + y² = 1.

Therefore,

cos²θ+sin²θ=1.

2. The center is at x=2, y=-3. The radius is √81 = 9.

3. Is this a trig equation? It is equivalent to:

2(1-sin²θ)=sinθ+1 <=>

2 – 2sin²θ = sinθ+1 <=>

2sin²θ + sinθ -1 = 0

Denote x = sinθ. You get a quadratic equation in x:

2x² + x -1 = 0

The solutions are

x=1/2

x=-1

So, you have:

sinθ=1/2

sinθ=-1

The answer is:

θ=(-1)^n π/6 + nπ, n is any integer

θ=-π/2 + 2nπ, n is any integer

1.

This is Pythagoras’ Theorem in trigonometric form. As it is usually the first trig formula anyone learns, you should prove it by drawing a right-angled triangle.

Call it ABC, let B be the right angle, and x be the angle at A.

Then:

AC^2 = AB^2 + BC^2

Divide by AC^2:

1 = AB^2 / AC^2 + BC^2 / AC^2

1 = (AB / AC)^2 + (BC / AC)^2

cos^2(x) + sin^2(x) = 1.

2.

That is a circle of radius 9 centred at (2, – 3).

The equation of a circle of radius r centred at (h, k) is:

(x – h)^2 + (y – k)^2 = r^2.

3.

As an identity, this formula is incorrect.

If you put $ = 0, the LHS = 2 and the RHS = 1.

4.

You are looking for terms a, ar, ar^2, ar^3, ar^4 with:

a = 4

ar^4 = 1024 ……(1)

Dividing (1) by 4:

r^4 = 256

r = 4

Hence the three geometric means are:

ar, ar^2, ar^3

= 16, 64, 256.

5.

I presume you want to factorise this.

2k^2 + 3k – 5

= (2k + 5)(k – 1).

6.

If T is the first term and T3 the third term, then the sequence is:

26, 26+d, 26+2d, 26+3d ….. with k-th term 26 + (k – 1)d.

As T3 = 10:

26 + 2d = 10

2d = -16

d = -8.

The k-th term is therefore:

26 + (k – 1)(-8)

= 34 – 8k.

Draw a right triangle triangle ABC with angle ABC as 90*

in the above figure,

sin^C=AB/AC cos^C=BC/AC

hence, sin^2 C + cos^2 C= AB^2/AC^2 +BC^2/AC^2

=(AB^2+BC^2)/AC^2

but according to Pythagoras theory, AB^2+BC^2=AC^2

so sin^2 C + cos^2 C =AC^2/AC^2 = 1