We will learn step-by-step the proof of compound angle formula sin (α + β). Here we will derive formula for trigonometric function of the sum of two real numbers or angles and their related result. The basic results are called trigonometric identities.
The expansion of sin (α + β) is generally called addition formulae. In the geometrical proof of the addition formulae we are assuming that α, β and (α + β) are positive acute angles. But these formulae are true for any positive or negative values of α and β.
Now we will prove that, sin (α + β) = sin α cos β + cos α sin β; where α and β are positive acute angles and α + β < 90°.
Let a rotating line OX rotate about O in the anti-clockwise direction. From starting position to its initial position OX makes out an acute ∠XOY = α.
Again, the rotating line rotates further in the same
direction and starting from the position OY makes out an acute ∠YOZ
= β.
Thus, ∠XOZ = α + β < 90°.
We are suppose to prove that, sin (α + β) = sin α cos β + cos α sin β.
Construction: On the bounding line of the compound angle (α + β) take a point A on OZ, and draw AB and AC perpendiculars to OX and OY respectively. Again, from C draw perpendiculars CD and CE upon OX and AB respectively. |
Proof: From triangle ACE we get, ∠EAC = 90° - ∠ACE = ∠ECO = alternate ∠COX = α.
Now, from the right-angled triangle AOB we get,
sin (α + β) = \(\frac{AB}{OA}\)
= \(\frac{AE + EB}{OA}\)
= \(\frac{AE}{OA}\) + \(\frac{EB}{OA}\)
= \(\frac{AE}{OA}\) + \(\frac{CD}{OA}\)
= \(\frac{AE}{AC}\) ∙ \(\frac{AC}{OA}\) + \(\frac{CD}{OC}\) ∙ \(\frac{OC}{OA}\)
= cos ∠EAC sin β + sin α cos β
= sin α cos β + cos α sin β, (since we know, ∠EAC = α)
Therefore, sin (α + β) = sin α cos β + cos α sin β. Proved.
1. Using the t-ratios of 30° and 45°, evaluate sin 75°
Solution:
sin 75°
= sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30
= \(\frac{1}{√2}\) ∙ \(\frac{√3}{2}\) + \(\frac{1}{√2}\) ∙ \(\frac{1}{2}\)
= \(\frac{√3 + 1}{2√2}\)
2. From the formula of sin (α + β) deduce the formulae of cos (α + β) and cos (α - β).
Solution:
We know that, sin (α + β) = sin α cos β + cos α sin β …….. (i)
Replacing α by (90° + α) on both sides of (i) we get,
sin (90° + α + β)
= sin {(90° + α) + β} = sin (90° + α) cos β + cos (90° + α) sin β, [Applying the formula of sin (α + β)]
⇒ sin {90° + (α + β)} = cos α cos β - sin α sin β, [since sin (90° + α) = cos α and cos (90° + α) = - sin α]
⇒ cos (α + β) = cos α cos β - sin α sin β …….. (ii)
Again, replacing β by (- β) on both sides of (ii) we get,
cos (α - β) = cos α cos (- β) - sin α sin (- β)
⇒ cos (α - β) = cos α cos β + sin α sin β, [since cos (- β) = cos β and sin (- β) = - sin β]
3. If sin x = \(\frac{3}{5}\), cos y = -\(\frac{12}{13}\) and x, y both lie in the second quadrant, find the value of sin (x + y).
Solution:
Given, sin x = \(\frac{3}{5}\), cos y = -\(\frac{12}{13}\) and x, y both lie in the second quadrant.
We know that cos\(^{2}\) x = 1 - sin\(^{2}\) x = 1 - (\(\frac{3}{5}\))\(^{2}\) = 1 - \(\frac{9}{25}\) = \(\frac{16}{25}\)
⇒ cos x = ± \(\frac{4}{5}\).
Since x lies in the second quadrant, cos x is – ve
Therefore, cos x = -\(\frac{4}{5}\).
Also, sin\(^{2}\) y = 1 - cos\(^{2}\) y = 1 - (-\(\frac{12}{13}\))\(^{2}\) = 1 - \(\frac{144}{169}\) = \(\frac{25}{169}\)
⇒ sin y = ± \(\frac{5}{13}\)
Since y lies in the second quadrant, sin y is + ve
Therefore, sin y = \(\frac{5}{13}\)
Now, sin (x + y) = sin x cos y + cos x sin y
= \(\frac{3}{5}\) ∙ (- \(\frac{12}{13}\)) + (- \(\frac{4}{5}\)) ∙ \(\frac{5}{13}\)
= - \(\frac{36}{65}\) - \(\frac{20}{65}\)
= - \(\frac{56}{65}\)
4. If m sin (α + x) = n sin (α + y), show that, tan α = \(\frac{n sin y - m sin x}{m cos x - n cos y}\)
Solution:
Given, m sin (α + x) = n sin (α + y)
Therefore, m (sin α cos x + cos α sin x) = n (sin α cos y+ cos α sin y), [Applying the formula of sin (α + β)]
m sin α cos x + m cos α sin x = n sin α cos y + n cos α sin y,
or, m sin α cos x - n sin α cos y = n cos α sin y - m cos α sin x
or, sin α (m cos x - n cos y) = cos α (n sin y - m sin x)
or, \(\frac{sin α}{cos α}\) = \(\frac{n sin y - m sin x}{m cos x - n cos y}\).
or, tan α = \(\frac{n sin y - m sin x}{m cos x - n cos y}\). Proved.
11 and 12 Grade Math
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