Method II
Given equation sin q = k. Let a be the least positive angle, such that sin a = k. Also, any angle co-terminal with the angles a , p - a is trigonometrically equivalent and so has the same 'sine'. ..
Given equation sin q = k. Let a be the least positive angle, such that sin a = k. Also, any angle co-terminal with the angles a , p - a is trigonometrically equivalent and so has the same 'sine'. ..Summary
An alternating current is that which changes continuously in magnitude and periodically in direction. It can be represented by a sine curve or a cosine curve i.e., I = I o sin w t or I = I o = cos w t Here, I o is peak value of current and I is instantaneous value of current. w = ..
Trigonometry
to q is the side AB. The hypotenuse of the D ABC is the side AC. Now, let us write down the trigonometrical ratios with the help of the above triangles: (i) Sine q : It is defined as the ratio of the side opposite to q and the hypotenuse. i.e., In short we write sin (ii) Cosine q : It is ..
to q is the side AB. The hypotenuse of the D ABC is the side AC. Now, let us write down the trigonometrical ratios with the help of the above triangles: (i) Sine q : It is defined as the ratio of the side opposite to q and the hypotenuse. i.e., In short we write sin (ii) Cosine q : It is ..Some Trigonometrical Identities
1. sin A = cos (90 o - A) 2. 3. tan A x tan (90 o - A) = 1 4. sin 2 A + cos 2 A = 1 5. 1 + tan 2 A = sec 2 A 6. 1 + cot 2 A = cosec 2 A Let us prove the above identities. Let D ABC be a right-angled triangle with B = 90 o . Let BC = a, AC = b and AB = c. (1..
1. sin A = cos (90 o - A) 2. 3. tan A x tan (90 o - A) = 1 4. sin 2 A + cos 2 A = 1 5. 1 + tan 2 A = sec 2 A 6. 1 + cot 2 A = cosec 2 A Let us prove the above identities. Let D ABC be a right-angled triangle with B = 90 o . Let BC = a, AC = b and AB = c. (1..See what our Users say :
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