This article covers some concepts that are taught in classes 11th and 12th, and hence new for arts/commerce candidates.
Few things which the above figure represents :
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- There are 4 quadrants (shown with I, II, III and IV).
- Quadrant I – 0° to 90°
- Quadrant II – 90° to 180°
- Quadrant III – 180° to 270°
- Quadrant IV – 270° to 360°
- There are 4 quadrants (shown with I, II, III and IV).
- In the first quadrant, all the trigonometric functions are positive. So the values of sin56, cos18, tan89, cot67, cosec33, etc. are positive. Note that I have taken the angles 56, 18, 89, 67, 33 and all of them are less than 90(hence belong to the first quadrant)
- In the second quadrant, only sin and cosec are positive, and rest are negative. Hence sin91, sin135, cosec120, sin116, etc. are positive while cos135, cot120, tan95, etc, are negative.
- In the third quadrant, only tan and cot are positive, and rest are negative. Hence tan200, tan198, cot255, etc. are positive while cos255, sin220, cosec265, etc, are negative.
- In the fourth quadrant, only cos and sec are positive, and rest are negative. Hence cos300, cos350, sec290, sec285, etc. are positive while tan359, cot355, sin340, etc, are negative.
- The mnemonic to remember which trigonometric function is positive in which quadrant is – All Students Take Calculus. “All” is the first word of the sentence and hence represents the first quadrant. All trigonometric functions are positive in the 1st quadrant. Second initial is “S” which represents sin (indicating sin/cosec are positive in 2nd quadrant). Third initial is “T” which represents tan (indicating tan/cot are positive in 3rd quadrant). Fourth initial is “C” which represents cos (indicating cos/sec are positive in 4th quadrant).
- All trigonometric functions are positive in the 1st quadrant
- sin/cosec are positive in 2nd quadrant (sin and cosec are reciprocal of each other and hence their signs are same)
- tan/cot are positive in 3rd quadrant
- cos/sec are positive in 4th quadrant
Converting trigonometric functions :
sin(90 – A) = cosA, and hence sin65 = sin(90 – 25) = cos25
sin(90 + A) = cosA, and hence sin135 = sin(90 + 45) = cos45
cos(90 – A) = sinA, and hence cos85 = cos(90 – 5) = sin5
cos(90 + A) = -sinA, and hence cos135 = cos(90 + 45) = -sin45 = -(1/√2)
where A is any acute angle
- sin(90 + A) refers to a value in the 2nd quadrant because A is an acute angle and hence (90 + A) would cover angles from 90 to 180 degrees (depending upon the value of A). 90 to 180 degrees is the range of 2nd quadrant. Now we have seen that in the second quadrant sin is positive. Hence sin(90 + A) = +cosA.
- cos(90 + A) = -sinA, because in the second quadrant cos is negative.
- (90 – A) represents the 1st quadrant and in the 1st quadrant, all the trigonometric functions are positive, hence:
- sin(90 – A) = +cosA and hence sin75 = sin(90 – 15) = cos15 (here A = 15)
- cos(90 – A) = +sinA
- tan(90 – A) = +cotA
- cot(90 – A) = +tanA
- sec(90 – A) = +cosecA
- cosec(90 – A) = +secA
- Now instead of 90 degrees if we have 180 degrees, then the functions are not converted. E.g.
- sin(180 – A) = sinA, and hence sin135 = sin(180 – 45) = sin45
- cos(180 – A) = -cosA and hence cos165 = cos(180 – 15) = -cos15
- tan(180 – A) = -tanA
- cosec(180 – A) = cosecA
- sec(180 – A) = -secA
- Note that in the above lines only sin and cosec are positive, because (180 – A) represents 2nd quadrant and in the second quadrant only sin and cosec are positive.
- sin is converted into cos
- tan is converted into cot
- sec is converted into cosec
You need only this much knowledge to solve SSC questions [You don’t have to do PhD after all :)]
Now let us solve some CGL questions:
We have to convert sin3A into cos.
3A is an acute angle and hence sin3A lies in the 1st quadrant.
sin3A = +cos(90 – 3A) [sin is positive in 1st quadrant, hence we have written +cos(90 – 3A)]
Hence, cos(90 – 3A) = cos(A – 26)
90 – 3A = A – 26 [Equating cos]
4A = 116
or A = 29
Note: In this question we had to convert sin into cos, hence we checked the sign of sin.
cos20 = cos(90 – 70) = +sin70 [cos20 lies in the 1st quadrant and cos is positive in the 1st quadrant, hence we have written +sin70]
Hence, sin5θ = sin70
5θ = 70 [Equating sin]
θ = 14
Answer : (D)
Note: In this question, don’t write sin5θ = cos(90 – 5θ), because this formula is applicable only for acute angles and 5θ is not necessarily an acute angle
I hope this concept of quadrants and conversion is clear.
Moving on, I discussed the basic trick of Trigonometry (putting the value of theta) in Part-1. Since this trick is extremely important, in each article of trigonometry I will solve some questions by assuming the value of θ so that you imbibe that method well. For reference, I am attaching the values-
Let us take some more questions from CGL-
In this question you cant take θ = 45, 90 because that will make 2cos^4θ – cos^2θ = 0, and we know that denominator can never be zero. So you are left with two values: 0, 30 or 60
Take θ = 0
sec0 = 1
sin0 = 0
cos0 = 1
Put the above values and you will get the value of the expression as 1.
Put A = 30
The value of the expression = 1/2/(1 + √3/2) + 1/2(1 – √3/2) = (2 – √3) + (2 + √3) = 4
Now put A = 30 in all the 4 options
(A) 4 (B) 4/√3 (C) 1 (D) √3
Put θ = 45
a = 0, b = 0
Value of the expression = (0 + 4)(0 – 1)^2 = 4
Answer : (A)
α = 30 and β = 60