# Math Assignment Help With Tangent Function

## 9.7 Tangent function

The figure is a unit circle, with origin O as center cuts the *x*-axis at A (1, 0) and let a variable point moving on the circumference move through an arc length q. i.e., AP = p (q). The coordinates at the position of p (q) are p(*x*, y) = (cos q, sin q).

Then the tangent function is defined in the form as

Tanθ = y/*x *where *x *≠ 0

Tanθ = sinθ/ cosθ, where cosθ ≠ 0

X = 0 or cosθ = 0 for θ = (nπ + π/2), where n is an integer. Therefore tanθ is defined for all θ€R except

θ = (nπ + π/2), n€ Z

Therefore, the domain of the tanθ is R’ = R – (nπ+π/2); n€ Z

**Theorem1.** Sin^{2}θ + Cos^{2}θ = 1

Proof: Let X’OX and Y’OY be the coordinate axes and O being the center. Draw a circle of unit radius cutting OX at A. Let a moving point starts from A, moving along the circumference of the circle. Let its final position be P (*x*, y), and arc AP=θ.

We know that the equation of the circle is *x*^{2} + y^{2} = 1

Since the point P (Cosθ, Sinθ) lies on it, therefore

Cos^{2}θ + Sin^{2}θ = 1

**Theorem3.** Let XOX' and YOY' be the rectangular coordinate axes. Taking O as center and radius = 1, draw a circle to cut *x*-axis at A and A' and y-axis at B and B'. Let the rotating line start from OA revolving in anticlockwise direction and taking the final position p (*x*,y) so that arc AP = θ.

On the other hand, if the point starts from A and moves in clockwise direction through arc length AP’ equal arc length AP, then

AP’ = -θ

therefore,

Join P and P'. Triangles POM and P'OM are congruent.

\ PM = MP'. If PM = y then P'M = -y.

From the DPOM, sin q = y and cos q = x

In D P'OM, sin (-q) = -y and cos (-q) = x

sin(-θ) = -sin θ

cos(-θ) = cos θ

**Theorem4.** Let *x* and y be any two real numbers and let P(*x*)and Q(y) be the corresponding trigonometric points on the unit circle. In the above figures we have taken *x* and y so that

π/2 < y< *x *<π

The coordinates of P(*x*) are (cos*x*, sin *x*) and that of Q(y) are (cos y, sin y) by the definitions of cosine and sine functions.

Now choose a point R on the unit circle so that arc AR has a measure of (x-y) units. Then the trigonometric point with respect to (*x-*y) is R(*x*-y) and the corresponding coordinates of R are (cos(*x*-y),sin(*x*-y)). The arc length of AR is the same as arc PQ and hence the chord lengths of PQ and AR are same.

- i) |AR| = distance from R to A.

Using distance formula

[d^{2} = (*x*1 – *x*2)^{2} + (y1 – y2)^{2}]

|AR|^{2} = [cos(*x* – y) – 1]^{2}+ [sin(*x *– y) – 0]^{2}

= cos^{2}(*x* – y) – 2 cos(*x *– y) + 1 + sin^{2} (*x* – y)

= [cos^{2} (*x* – y) + sin^{2} (*x* – y)] – 2 cos(*x* – y) + 1

= 1 – 2 cos(*x* – y) +1

= 2 – 2cos(*x* – y) …..(i)

|PQ|^{2} = (cos*x *– cosy)^{2} + (sin*x *– siny)^{2}

= cos^{2}*x* – 2cos*x*cosy+cos^{2}y +sin^{2}*x* +2sin*x*siny + sin^{2}y

= (cos^{2}*x* + sin^{2}*x*) + (cos^{2}y + sin^{2}y) – 2[cos*x*cosy + sin*x*siny]

= 1 + 1 - 2[cos*x*cosy + sin*x*siny]

= 2 - 2[cos*x*cosy + sin*x*siny] ….(ii)

Since AR = PQ, therefore,

2 – 2cos(*x* – y) = 2 -2[cos*x*cosy + sin*x*siny]

-2 cos(*x* – y) = -2[cos*x*cosy + sin*x*siny]

**cos( x – y) = cosxcosy +sinxsiny**

**ii)****cos(***x*+y)

** = cos[x – (-y)]**

** = (cos x) {cos(-y)} + (sinx) {sin(-y)}**

** = cos xcosy + [-sinxsiny] [since cos(-x) = cosx & sin(-y) = -siny]**

** = ****cos xcosy - sinxsiny**

**Theorem5. Proof: i) cos( π/2 – x) = sinx**

cos (π/2 – *x*) = cos (π/2)cos*x *+ sin(π/2)sin*x*

= 0 +(1) sin*x*

= **sin x**

**ii)****sin(****π/2 –***x*) = cos*x*

sin(π/2 – *x*) = cos[π/2 – (π/2 –* x*)] [since cos(π/2 – θ) = sinθ]

= **cos x**

**iii) cos(π/2 +***x*) = -sin*x*

cos(π/2 + *x*) = cos π/2cos*x* - sin π/2sin*x*

= 0 – (1) sin*x*

= **- sin x**

**iv)****sin(****π/2 +***x*) = cos*x*

sin(π/2 + *x*) = sin[π/2 – (-*x*)]

= cos(-*x*)

**= cos x**

**Theorem 8**

For all real values of *x*.

**i) cos(π –***x*) = -cos*x*

**Proof:** cos(π – *x*) = cosπ cos*x* + sinπ sin*x*

= (-1)cos*x* + 0 sin*x*

= **- cos x**

**ii)****sin(π –***x*) = sin*x*

**Proof:** sin(π – *x*) = sin[(π/2) + (π/2) – *x*]

= sin[(π/2) + (π/2 – *x*)]

= cos(π/2 – *x*)

**= sin x**

**iii) cos(π+***x*) = -cos*x*

**Proof:** cos(π+*x*) = cosπcos*x* – sinπsin*x*

= (-1)cos*x* – (0)sin*x*

**= -cos x**

**iv) sin(π +***x*) = -sin*x*

Proof: sin(π + x) = sin [(π/2) + (π/2 + *x*)]

= cos(π/2 +*x*)

= **- sin x**

**Theorem 9**

For real values of *x*

**i) cos(2π+***x*) = cos*x*

**Proof:** cos(2π+*x*) = cos2πcos*x* – sin2πsin*x*

= (1)cos*x* – (0)sin*x*

= **cos x**

**ii)****sin(2π+***x*) = sin*x*

**Proof :** sin(2π+ *x*) = sin [π + (π + *x*)]

= - [-sin(π+*x*)]

= - [-sin*x*]

= **sin x**

**Theorem 10**

**i)****sin(***x*+y) =sin*x*cosy + cos*x*siny

**Proof: **sin( x+y) = cos[π/2 – (*x* +y)]

= cos[(π/2 – *x*) – y]

= cos(π/2 – *x*) cosy + sin(π/2 – *x*) siny

= **sin xcosy + cosxsiny**

**ii) sin(***x*– y) = sin*x*cosy - cos*x*siny

**Proof: **sin(*x* – y) = sin[*x* + (-y)]

= sin*x*cos(-y) + cos*x*sin(-y)

= **sin xcosy – cosxsiny **

**Theorem 11**

**i)****sin2***x*= 2sin*x*cos*x*

**Proof:** Consider

sin(*x*+y) = sin*x*cosy + cos*x*siny

Now put *x* = y, then

Sin(*x* + *x*) = sin*x*cos*x* + cos*x*sin*x*

**Sin2 x = 2 sinx cosx **

**ii) cos2***x*= cos^{2}*x*– sin^{2}*x*= 2cos^{2}*x*– 1 = 1 – 2sin^{2}*x*

**Proof;** take,

cos(*x*+y) = cos*x*cosy – sin*x*siny

now place x = y,

cos(*x*+*x*) = cos*x*cos*x* – sin*x*sin*x*

= cos^{2}*x* – sin^{2}*x*

= cos^{2}*x* – (1 - cos^{2}*x*)

= 2cos^{2}*x* – 1

= 2(1 – sin^{2}*x*) – 1

= **1 - 2 sin ^{2}x**

**iii) sin3***x*= 3sin*x*– 4 sin^{3}*x*

**Proof:**we can write,

Sin3*x* = sin(2*x* + *x*)

= sin2*x*cos*x* + cos2*x*sin*x*

= (2sin*x*cos*x*) cos*x* + (1 – 2sin^{2}*x*) sin*x*

= 2sin*x* cos^{2}*x* + sin*x* – 2 sin^{3}*x*

= 2sin*x* – 2sin^{3}*x* + sin*x* – 2 sin^{3}*x*

= **3sin x – 4sin^{3}x**

**cos3**

*x*= 4 cos^{3}*x*– 3cos*x***Proof: ** cos3*x* = cos(2*x* + *x*)

= cos2*x*cos*x* – sin2*x*sin*x*

= (2cos^{2}*x* – 1) cos*x* – 2sin*x*cos*x*sin*x*

= 2 cos^{3}*x* – cos*x* – 2cos*x*sin^{2}*x*

= 2 cos^{3}*x* – cos*x* – 2cos*x*(1 - cos^{2}*x*)

= 2cos^{3}*x* – cos*x* – 2cos*x* + 2cos^{3}*x*

**= 4cos ^{3}x – 3cosx**

**Theorem 12**

**i) 2sin***x*cosy = sin(*x*+ y) + sin(*x*– y)**ii) 2cos***x*siny = sin(*x*+ y) – sin(*x*– y)**iii) 2cos***x*cosy = cos(*x*+ y) + cos(*x*– y)**iv) 2sin***x*siny = cos(*x*– y) – cos(*x*+ y)

**Proof:**

The domain of the tangent function is R’ = R - nπ +π/2 : nϵ Z

Adding (1) and (2)

**sin( x+y) + sin(x – y) = 2 sinx cosy ….(i)**

Subtracting 2 from 1

**sin( x+y) - sin(x – y) = 2 cosx siny ….(ii)**

Consider cos(*x*+y) = cos*x* cosy – sin*x* siny …..(3)

And,

**cos( x – y) = cosxcosy +sinxsiny ….(4)**

Adding (3) & (4)

** cos( x + y) + cos(x – y) = 2cosxcosy ….(iii)**

Subtracting (4) from (3)

cos(*x* + y) – cos(*x* – y) = - 2sin*x*siny

Or

**cos( x – y) – cos(x + y) = 2sinxsiny …(iv)**

**Theorem 13**

- i) 1 – cos
*x*= 2sin^{2}(*x*/2)

**Proof:** Consider

cosx = cos(*x*/2 + *x*/2) = cos [2(*x*/2)]

= cos^{2} (*x*/2) – sin^{2}(*x*/2)

= 1 – sin^{2}(*x*/2) – sin^{2}(*x*/2)

cos*x* = 1 – 2sin^{2}(*x*/2)

**1**

**– cosx = 2sin**

^{2}(*x*/2)**ii) 1 + cosx = 2cos**^{2}(x/2)

Proof: cos*x* = cos(*x*/2 + *x*/2) = 2cos[2(*x*/2)]

= cos^{2} (*x*/2) – sin^{2}(*x*/2)

= cos^{2}(*x*/2) – [1 – cos^{2}(*x*/2)]

cosx = 2 cos^{2}(*x*/2) – 1

** 1 + cosx = 2 cos ^{2}(x/2)**

**Theorem 14**

**i) sin x + siny/2 = 2 sin (x + y) cos(x – y)**/2

**ii) sin x + siny/2 = 2 cos (x + y) sin(x – y)**/2

**iii) cos x + cosy/2 = 2 cos(x+y)sin(x – y)**/2

**iv) cos x - cosy/2 = - 2 sin(x+y)sin(x – y)**/2

Proof:

Sin(A+B) = sinAcosB +cosAsinB ….(a)

Sin(A – B) = sinAcosB – cosAsinB ….(b)

Adding (a) and(b)

Sin(A+B) +sin(A – B) = 2sinAcosB

Let A+B = *x* and A – B = y

Then,

A = *x* + y/2 and B = *x* – y/2

Therefore,

**sin x + siny/2 = 2 sin (x + y) cos(x – y)/2 ….(i)**

Subtracting (b) from (a)

sin(A + B) – sin(A – B) = 2cosAsinB

sinx + siny/2 = 2 cos (x + y) sin(x – y)/2 ….(ii)

Now take,

cos(A + B) = cosA cosB – sinA sinB ….(c)

cos(A – B) = cosA cosB + sinA sinB …(d)

adding ( c) and (d)

cos(A + B) + cos(A – B) = 2 cosA cosB

cosx + cosy/2 = 2 cos(x+y)sin(x – y)/2 ….(iii)

Subtracting (d) from ( c)

cos(A + B) – cos(A – B) = -2sinAsinB

**cos x - cosy/2= - 2 sin(x+y)sin(x – y)/2 ….(iv)**

**Theorem 15**

**i) Sin( x +y) sin(x – y) = sin^{2}x – sin^{2}y**

**Proof: **

**sin( x + y)sin(x – y)**

= (sin*x* cosy + siny cos*x*)(sin*x* cosy – cos*x* siny)

= sin^{2}*x* cos^{2}y – sin^{2}y cos^{2}*x*

= sin^{2}*x* (1 – sin^{2}y) – sin^{2}y(1 – sin^{2}*x*)

= sin^{2}*x* – sin^{2}*x* sin^{2}y – sin^{2}y + sin^{2}*x* sin^{2}y

= **sin ^{2}x – sin^{2}y**

**ii) cos(***x*+ y) cos(*x*– y) = cos^{2}*x*– cos^{2}y

**Proof:**

**cos( x+y) cos(x – y)**

= (cos*x* cosy – sin*x* siny) (cos*x* cosy + sin*x* siny)

= cos^{2}*x* cos^{2}y – sin^{2}*x* sin^{2}y

= cos^{2}*x* (1 – sin^{2}y) – sin^{2}y(1 – cos^{2}*x*)

= cos^{2}*x* – cos^{2}*x* sin^{2}y – sin^{2}y + sin^{2}y cos^{2}*x*

= **cos ^{2}x – cos^{2}y**

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