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As in the last section we have learned about the complementary angles and heights and distances, in this section we will learn about Trigonometry (Value Based and Simplification). We learned the values of 30o, 45o, 60o, 90o of different trigonometric functions, here will see how we find out the values of the other trigonometric values.

Signs of Trigonometric Ratios

We have already discussed the signs of trigonometric ratios. Let’s look one more time.

QUADRANTSinθCosθTanθCosecθSecθCotθ
( 0o, 90o) Ist quad.     +     +     +     +     +     +
( 90o, 180o) IInd quad.     +     –     –     –     –     +
(180o, 270o) IIIrd quad.     –     –     +     +     –     –
(270o, 360o) IVth quad.     –     +     –     –     +     –

You are very friendly with it as we have discussed it earlier.

The main criteria for me to discuss this is to introduce you to the trigonometric ratios of the allied angles.

Trigonometric ratios of allied angles:-

Two angles are said to be allied when their sum or difference is either zero or a multiple of 90o. The angles θ, 90o ± θ, 180o ± θ, 270o ± θ, 360o – θ, etc. are called allied to the angle θ if we measure θ in degrees.

Look at the table below, you will have a clear picture of what the trigonometric functions look like when we look at the allied angles.

     θ    Sin θCosec θCos θ   Sec θ     Tan θ  Cot θ
   – θ– Sin θ– Cosec θ Cos θ  Sec θ     -Tan θ   -Cot θ
90o – θ Cos θ Sec θ    Sin θ  Cosec θ   Cot θ   Tan θ
90o ­­+ θ Cos θ Sec θ  – Sin θ -Cosec θ   -Cot θ   -Tan θ
180o – θ Sin θ  Cosec θ -Cos θ  -Sec θ     -Tan θ   -Cot θ
180o + θ -Sin θ– Cosec θ -Cos θ  -Sec θ     Tan θ   Cot θ
270o – θ -Cos θ  -Sec θ  – Sin θ -Cosec θ   Cot θ    Tan θ
270o + θ -Cos θ -Sec θ  – Sin θ  Cosec θ   -Cot θ   -Tan θ
360o – θ -Sin θ– Cosec θ  Sin θ  Sec θ     -Tan θ   -Cot θ

As you see, what actually happens is that the function doesn’t change when we are in (– θ), 180o ± θ, 360o – θ, the only change we see is the sign change. While in the rest of the scenarios sin gets converted in cos, cosec into sec and tan into cot and vice – versa.

NOTE:- This table is very important throughout your prep. Also, this covers the basic simplification questions.

NOTE:- 180o = πc { where radian is denoted by c}

Let me give you an example regarding this table.

See, if you wanted to find the value of say, sin150o.

You don’t know the value of this, no one does if they don’t know about the allied angles.

Look carefully, sin150o is close to sin180o

We can write sin150o = sin(180o – 30o) = sin30o = ½  {sin(180 – θ)= sinθ}

As this must be clear to you now let’s move ahead with some simplification questions asked in SSC-CGL EXAM. Let’s look at them.

Ques:- Simplify:- 4/3 cotπ/6 + 3 cos2 150o – 4 cosec2 45o + 8 sin π/2             

Solution:-  If you look at the question carefully you see you need not do the tedious calculation.

Before solving this question broaden your thinking and make this equation simpler by first observing what values it is asking.

1. You know here, π = 180o

2. The values of cos150o and cosec 45o.

As we have already learned that we can write,

 cos150o = cos( 180o – 30o) = – cos30o = – 3  /2     {cos(180o – θ) = cosθ}

 See you did it… Hurray!! You’re a champ now 😉

When it comes to the value of cosec 45o you already know what it is i.e., cosec 45o = 2  

SO, we have everything sorted let’s solve it now.

4/3 cotπ/6 + 3 cos2 150o – 4 cosec2 45o + 8 sin π/2

= 4/3 cot2 180/6 + 3(-3  /2)2 – 4 (2) 2 + 8 sin180/2

= 4/3 cot2 30o + 3 ×3/4 – 4× 2 + 8 sin90o

=4/3 (3) 2 + 9/4 – 8 + 8 ×1

= 4 + 9/4 – 8 +8 = 25/4

See you did it !!! yippie!!

we are going to quickly look some of the formulae related to trigonometry and further, we will solve the questions based on them which we asked in the previous papers.

FORMULAE

Let us quickly look at some important formulae.

(1) sin (A+B) =sinA. cosB + cosA sinB

(2) sin(A – B) =sinA. cosB – cosA sinB

(3) cos(A+B) =cosA. cosB – sinA sinB

(4) cos(A-B) = cosA. cosB + sinA sinB

(5) 2 sinA.cosB = sin(A+B) + sin (A-B)

(6) 2 cosA. sinB = sin(A+B) – sin (A-B)

(7) 2 sinA. sinB = cos(A-B) – cos(A+B)

(8)  2 cosA.cosB = cos(A+B) + cos(A-B)

(9) sin²A – sin²B = sin(A+B). sin(A-B)

(10) cos²A – cos²B = cos(A+B).cos (A-B)

(11) tan(A + B) = tanA + tanB / 1 – tanA tanB

(12) tan(A – B) = tanA – tanB / 1 + tanA tanB

(13) cot(A + B) = cotA cotB – 1/cotA + cotB

(14) cot(A – B) = cotA cotB + 1/cotA – cotB

(15) sin2A = 2sinA. cosA

(16) cos2A = cos2A – sin2A

                   = 2cos2A – 1

                   = 1 – 2sin2 A

(17) tan2A = 2 tanA / 1 – tan2A

These were two angle formulae, we also have 3 angle formulae.

Let’s check them.

(18) sin3A = 3sinA – 4sin3A

(19) cos3A = 4cos3A – 3cosA

(20) tan3A = 3tanA – tan3A / 1 – 3tan2A

Ques:- What is the value of sin (B – C) cos (A – D) + sin (A – B) cos (C –D) + sin (C – A) cos (B – D)?

Solution: As we have already learned the formulas a few seconds before. Here, there is nothing new to it. But the calculation is lil bit lengthy but practice and you’ll do it in a wink of a second!

Let’s solve partly,

  • sin (B – C) cos (A – D)

= (sinB . cosC – cosB . sinC) (cosA . cosD + sinA . sinD)

= sinB . cosC . cosA . cosD + sinA . sinB . cosC . sinD – cosA . cosB . sinC .  cosD – sinA . sinC . sinD . cosB

  • sin(A – B). cos (C – D)

= (sinA . cosB – cosA sinB) (cosC . cosD + sinC . sinD)

= sinA . cosB . cosC . cosD + sinA . sinC . cosB . sinD – cosA . cosC sinB . cosD – cosA . sinB . sinC sinD

  • sin (C – A) . cos (B – D)

= (sinC . cosA – cosC . sinA) (cosB . cosD + sinB . sinD)

 = sinC . cosA . cosB . cosD + sinC cosA . sinB . sinD – cosC . sinA . cosB . cosD – sinB . sinD . cosC . sinA

There as no hard core calculation we just wrote the formula and expanded.

We need to calculate,

sin (B – C) cos (A – D) + sin (A – B) cos (C –D)  + sin (C – A) cos (B – D)

by adding all the above calculations, we get sum = 0

just practice will help you grapple with such solutions. But they were easy right?

Ques:- What is the value of-

              tan(π/4 + A) × tan(3π/4 + A)      

Solution:-

tan(π/4 + A) × tan(3π/4 + A)           { We will apply the formula of tan(A+B)}

tan π/4 + tanA / 1 – tan π/4. tanA  × tan 3π/4 + tanA / 1 – tan 3π/4. tanA

= 1 + tanA / 1 – tanA × (-1) + tanA / 1 – (-1) tanA

= -1

See, what a simple calculation it was.

Ques:- What is the value of sin (90° + 2A) [4 – cos2 (90o – 2A)]?

(1) 2 (cos3A – sin3A)

 (2) 2 (cos3A + sin3A)

(3) 4 (cos6A + sin6A)

 (4) 4 (cos6A – sin6A)

Solution:-

sin(90o + 2A). 4 – cos(90o– 2A)

 = cos 2A. (4 – sin22A)                               {sin(90o – θ) = cosθ and cos (90o – θ) = sinθ}

= (cos2A –sin2A) [4 –(2sinA.cosA)2]         {Applying the formula for cos2A and sin2A}

 = (cos2A – sin2A).4 [(1 – sin2A.cos2A)]    { taking 4 as common}

 = 4.(cos2A – sin2A) ((sin2A + cos2A) 2 – sin2Acos2A)

 = 4(cos2A – sin2A) (sin4A + cos4A + 2 sin2 Acos2A – sin2A.cos2A)

= 4 (cos2A – sin2A) (sin4A + cos4A + sin2A.cos2A) = 4 (cos6A – sin6A) So, the correct option is (4).

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