The area of mathematics relating to the study of trigonometric function in relation to measurement in triangles is known as trigonometry. Trigonometry is the branch of mathematics which literally consists of three words, 'Tri' referring to 'three', 'gona' meaning 'angle' and 'metry' which means 'measurement'. Hence, trigonometry refers to the measurement of triangles. We shall only deal with right angled triangles.
An angle that is taken into consideration before finding out perpendicular and base in a right angled triangle are known as the reference angle. The reference angles are usually denoted by Greek Alphabets. The side in front of the reference angle is called perpendicular and the remaining side is called base. Perpendicular and base are denoted by p and b respectively. So, in short p, b and h are called the elements of a right angled triangle.
In figure, let ∠C = θ be the reference angle. Then,
AB = Perpendicular (p)
BC = Base (b)
AC = Hypotenuse (h)
Similarly, if \(\angle\)A becomes the reference angle then,
AB = Base (b)
BC = Perpendicular (p)
AC = Hypotenuse (h)
Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles.Geometrically, these are identities involving certain functions of one or more angles. For examples,
tan x sin x + cos x = sec x
Solution: We will only use the fact that sin^{2} x + cos^{2} x = 1 for all values of x.
L.H.S = tan x sin x + cos x
= \(\frac{sin x}{cos x}\) × sin x + cos x = \(\frac{sin^2 x}{cos x}\) + cos x
= \(\frac{sin^2 x}{cos x}\) + cos x
= \(\frac{sin^2 x}{cos x}\) + \(\frac{cos^2 x}{cos x}\)
= \(\frac{sin^2 x + cos^2 x}{cos x}\)
= \(\frac{1}{cos x}\)
= R.H.S
∴ L.H.S = R.H.S proved.
Proving Trigonometric Identities
Showing the both sides of an identity equal to each other by using various techniques is known as proving trigonometric identity. For example,
sinθ × cotθ = cosθ
Here,
L.H.S = sinθ × cotθ
= sinθ × \(\frac{cosθ}{sinθ}\)
= cosθ
= R.H.S
∴ L.H.S = R.H.S proved.
Techniques for proving Trigonometric Identities
Solution
(sinθ + cosθ) (sinθ + cosθ)
= sin\(\theta\)(sinθ + cosθ) (sinθ + cosθ) + cos\(\theta\)(sinθ + cosθ) (sinθ + cosθ)
= sin^{2}\(\theta\) + sin\(\theta\) × cos\(\theta\) + cos\(\theta\) × sin\(\theta\) + cos^{2}\(\theta\)
= sin^{2}\(\theta\) + sin\(\theta\) × cos\(\theta\) + cos\(\theta\) × sin\(\theta\) + cos^{2}\(\theta\)
= sin^{2}\(\theta\) + 2 sin\(\theta\) × cos\(\theta\) + cos^{2}\(\theta\)
Solution
(1- tan\(\theta\)) (1+tan\(\theta\)) (1+tan^{2}\(\theta\)) (1+tan^{4}\(\theta\))
= (1 - tan^{2}\(\theta\))(1 + tan^{2}\(\theta\))(1 + tan^{4}\(\theta\))
= (1 - tan^{4}\(\theta\))(1 + tan^{4}\(\theta\))
1 - tan^{8}\(\theta\)
Solution
\(\frac{1}{1 + cosθ}\) + \(\frac{1}{1-cosθ}\)
= \(\frac{1}{1 + cosθ}\) + \(\frac{1}{1-cosθ}\)
= \(\frac{1-cosθ +1+cosθ}{(1-cosθ)(1-cosθ)}\)
= \(\frac{2}{1-cosθ}\) = \(\frac{2}{sin^2θ}\) = 2 cosec^{2}θ
Solution
= \(\frac{1}{secA - tanA}\) - \(\frac{1}{secA + tanA}\)
= \(\frac{secA + tanA - (secA - tanA)}{sec^2A - tan^2A}\)
= \(\frac{secA + tanA - secA + tanA}{1}\)
= 2 tanA
L.H.S = \(\frac{1 - sin\alpha}{cos\alpha}\)
= \(\frac{1 - sin\alpha}{cos\alpha}\) × \(\frac{1 + sin\alpha}{1 + sin\alpha}\)
= \(\frac{1 - sin\alpha}{cos\alpha(1 + sin\alpha)}\)
= \(\frac{cos\alpha}{(1 + sin\alpha)cos\alpha}\)
= \(\frac{cos\alpha}{1 +sin\alpha}\) = R.H.S
\(\therefore\) LHS = RHS proved.
ASK ANY QUESTION ON Trigonometry
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May 25, 2017
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Rojan Rasaily
CosA-sinA 1/cosA SinA-1=cotA cosecA
Mar 31, 2017
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Buddhiram kewat
Cosecπ 1/cosecπ-1=tan^2π/sec^2-1
Mar 24, 2017
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Rajan
CotA tanB/cotB tanA=cotAtanB
Mar 19, 2017
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