We will learn step-by-step the proof of compound angle formula cos (α - β). Here we will derive formula for trigonometric function of the difference of two real numbers or angles and their related result. The basic results are called trigonometric identities.
The expansion of cos (α - β) is generally called subtraction formulae. In the geometrical proof of the subtraction formulae we are assuming that α, β are positive acute angles and α > β. But these formulae are true for any positive or negative values of α and β.
Now we will prove that, cos (α - β) = cos α cos β + sin α sin β; where α and β are positive acute angles and α > β.
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.
Now, the rotating line rotates further in the clockwise
direction and starting from the position OY makes out an acute ∠YOZ
= β (which is < α).
Thus, ∠XOZ = α - β.
We are suppose to prove that, cos (α - β) = cos α cos β + sin 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 produced BA respectively. |
Proof: From triangle ACE we get, ∠EAC = 90° - ∠ACE = ∠YCE = corresponding ∠XOY = α.
Now, from the right-angled triangle AOB we get,
cos (α - β) = \(\frac{OB}{OA}\)
= \(\frac{OD + DB}{OA}\)
= \(\frac{OD}{OA}\) + \(\frac{DB}{OA}\)
= \(\frac{OD}{OA}\) + \(\frac{CE}{OA}\)
= \(\frac{OD}{OC}\) ∙ \(\frac{OC}{OA}\) + \(\frac{CE}{AC}\) ∙ \(\frac{AC}{OA}\)
= cos α cos β + sin ∠CAE sin β
= cos α cos β + sin α sin β, (since we know, ∠CAE = α)
Therefore, cos (α - β) = cos α cos β + sin α sin β. Proved
1. Using the t-ratios of 30° and 45°, find the values of cos 15°.
Solution:
cos 15°
= cos (45° - 30°)
= cos 45° cos 30° - sin 45° sin 30°
= (\(\frac{1}{√2}\) ∙ \(\frac{√3}{2}\)) + (\(\frac{1}{√2}\) ∙ \(\frac{1}{2}\))
= \(\frac{√3 + 1}{2√2}\)
2. Prove the identities: sin 63°32’ sin 33°32’ + sin 26°28’ sin 56°28 = √3/2
Solution:
L. H. S. = Sin 63°32’ Sin 33°32’ + sin 26°28’ sin 56°28’
= sin(90° - 26° 28’) sin (90° - 56° 28’) + sin 26°28’ sin 56°28’
= cos 26°28’ cos 56°28’ + sin 26°28’ sin 56°28’
= cos (56°28’ - 26°28’)
= cos 30°
= \(\frac{√3}{2}\). Proved
3. Prove the identities:
1 + tan θ ∙ tan θ/2 = sec θ
Solution:
L.H.S = 1 + tan θ. tan θ/2
= 1 + \(\frac{sin θ ∙ sin θ/2}{cos θ ∙ cos θ/2}\)
= \(\frac{cos θ cos θ/2 + sin θ sin θ/2}{cos θ cos θ/2 }\)
= \(\frac{cos(θ - θ/2)}{cos θ cos θ/2}\)
= \(\frac{cos θ/2}{cos θ ∙ cos θ/2}\)
= \(\frac{1}{cos θ }\)
= sec θ. Proved
4. Prove that cos 70° cos 10° + sin 70° sin 10° = ½
Solution:
L.H.S. = cos 70° cos 10° + sin 70° sin 10°
= cos (70° - 10°)
= cos 60
= ½ = R.H.S. Proved
5. Find the maximum and minimum values of 3 cos θ + 4sin θ + 5.
Solution:
Let, r cos α = 3 …………… (i) and r sin α = 4 …………… (ii)
Now square the equation (i) and (ii) then add
r\(^{2}\) cos\(^{2}\) α + r\(^{2}\) sin\(^{2}\) α = 3\(^{2}\) + 4\(^{2}\)
⇒ r\(^{2}\) (cos\(^{2}\) α + sin\(^{2}\) α) = 25
⇒ r\(^{2}\) (1) = 25, since cos\(^{2}\) α + sin\(^{2}\) α = 1
⇒ r = 5, [Taking square root on both sides]
Now equation (i) divided by (ii) we get,
\(\frac{r sin α}{r cos α}\) = 4/3
⇒ tan α = 4/3
Therefore, 3 cos θ + 4 sin θ + 5 = r cos α cos θ + r sin α sin θ + 5
= 5 cos (θ - α) + 5
Since, -1 ≤ cos (θ - α) ≤ 1
Therefore, -5 ≤ 5 cos (θ - α) ≤ 5
⇒ -5 + 5 ≤ 5 cos (θ - α) + 5 ≤ 5 + 5
⇒ 0 ≤ 5 cos (θ - α) + 5 ≤ 10
From this inequality it readily follows that the maximum and minimum values of [5 cos (θ - α) + 5] i.e., (3 cos θ + 4 sin θ + 5) are 10 and 0 respectively.
6. Prove that sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos x
Solution:
L.H.S. = sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x
= cos (n + 2) x cos (n + 1) x + sin (n + 2) x sin (n + 1) x
= cos {(n + 2) x - (n + 1) x)
= cos x = R.H.S. Proved
11 and 12 Grade Math
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