Trigonometric Ratios of Multiple Angles

If A is an angle, then 2A, 3A, 4A, 5A, etc. are called multiple angles of A.

Summary

If A is an angle, then 2A, 3A, 4A, 5A, etc. are called multiple angles of A.

Things to Remember

sin2A = 2sinA. cosA tan2A =\(\frac{2cot A}{cot^2 A - 1}\)
sin2A =\(\frac{2tan A}{1 + tan^2 A}\) cot2A =\(\frac{cot^2 A - 1}{2cot A}\)
sin2A =\(\frac{2cot A}{1 + cot^2 A}\) cot2A =\(\frac{1 - tan^2 A}{2tan A}\)
cos2A = cos2A - sin2A 1 + cos2A = 2cos2A
cos2A = 1 - 2sin2A 1 - cos2A = 2 sin2A
cos2A = 2cos2A - 1 1 + sin2A = (cosA + sinA)2
cos2A =\(\frac{1 - tan^2 A}{1 + ttan^2 A}\) 1 - sin2A = (cosA - sinA)2
cos2A =\(\frac{cot^2 A - 1}{cot^2 A + 1}\) sin3A = 3 sinA - 4 sin \(^3\) A
tan2A =\(\frac{2 tan A}{1 - tan^2 A}\) cos3A = 4 cos\(^3\)A - 3cos A
tan3 A =\(\frac{3 tan A - tan^3 A}{1 - 3tan^2 A}\)  

 

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How was the Chinese society during the rule of Shang dynasty? Mention.


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Answer: <p>The Chinese society was strong and developed during the rule of Shang dynasty.</p>

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Explain the social condition during the reign of Zhou in China.


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Answer: <p>In the&nbsp;11<sup>th</sup> century, the Zhou (Chous) dynasty came to displace the Shang dynasty which started the feudal tradition. The Zhou used metal tools and weapons to build roads, expand irrigation canals to improve communication and commerce. This was also the classical age of Chinese Philosophy, with Confucianism, Daoism (Taoism), and Legalism all emerging during the Zhou dynasty. Nobles and the businessmen used to receive respect and privileges in this period.</p>

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What were the classes of people in China? Explain.


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Answer: <p>The classes of people in China were as follows:<br />There was a Monarchical system in China during the ancient times. Although the Chinese people had developed society, they didn&rsquo;t keep any written record until Shang dynasty of the 16<sup>th</sup> century. So, it was said that the Shang dynasty is the first civilization between 3000 to 3700 years ago. Art and culture developed during the Shang period. The people used to choose their strong king, who could protect them during the hard times. Slowly, the tradition of selecting king ended because the kings started to appoint their own son and relatives as the next kings. So, monarchy became hereditary.<br />In the&nbsp;11<sup>th</sup> century, the Zhou (Chous) dynasty came to displace the Shang dynasty which started the feudal tradition. The Zhou used metal tools and weapons to build roads, expand irrigation canals to improve communication and commerce. This was also the classical age of Chinese Philosophy, with Confucianism, Daoism (Taoism), and Legalism all emerging during the Zhou dynasty. Nobles and the businessmen used to receive respect and privileges in this period. <br />The King of Chin (Qin) seized control of China in 221 B.C. China got its present name from this dynasty. He extended territory and made the China a strong Kingdom. He built the famous &lsquo;The Great Wall of China&rsquo; in his period. It was the only manmade object visible from the moon. It has 8.5 meters high, 5.7 meters wide and 6000 meters long.</p>

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What are the main rivers of China?


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What are the large towns that are situated on the banks of Hwang Ho River?


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What  are the modern towns situated on the banks of Hwang-Ho river?


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Which river is known as Huang-Ho river?


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Trigonometric Ratios of Multiple Angles

Trigonometric Ratios of Multiple Angles

example for Trigonometric ratios of Multiple AnglesExample for Trigonometric ratios of Multiple Angles

If A is an angle, then 2A, 3A, 4A, 5A, etc. are called multiple angles of A. In this section we will discuss about the trigonometric ratios of angles 2A and 3A in terms of A.

(a) sin2A = sin(A + A) = sinA. cosA + cosA. sinA = 2sinA. cosA

(b) sin2A = 2sinA. cosA = \(\frac{2sinA. cosA}{1}\) =\(\frac{2sinA. cosA}{cos^{2}A + sin^{2}A}\) =\(\frac{\frac{2sinA. cosA}{cos^2A}}{\frac{cos^2A}{cos^2A} + \frac{sin^2A}{cos^2A}}\) = \(\frac{2tanA}{1 + tan^2A}\)

(c) sin2A =\(\frac{2tanA}{1 + tan^2 A}\) =\(\frac{\frac{2}{cotA}}{1 +\frac{1}{cot^2 A}}\) =\(\frac{2}{cotA}\)x\(\frac{cot^2A}{1 + cot^2 A}\) =\(\frac{2cotA}{1 + cot^2 A}\)

(d) cos2A = cos(A + A) = cosA. cosA - sinA. sinA = cos2A - sin2A

(e) cos2A = cos2A - sin2A = 1 - sin2A - sin2A = 1 - 2sin2A

(f) cos2A = cos2A - sin2A = cos2A - (1 - cos2A) = 2cos2A - 1

(g) cos2A = cos2A - sin2A =\(\frac{cos^2A - sin^2A}{1}\) =\(\frac{cos^2A - sin^2A}{cos^2A + sin^2A}\) =\(\frac{\frac{cos^2A}{cos^2A} - \frac{sin^2A}{cos^2A}}{\frac{cos^2A}{cos^2A} + \frac{sin^2A}{cos^2A}}\) =\(\frac{1 - tan^2A}{1 + tan^2A}\)

(h) cos2A =\(\frac{1 - tan^2A}{1 + tan^2A}\) =\(\frac{1 - \frac{1}{cot^2A}}{1 +\frac{1}{cot^2A}}\) =\(\frac{cot^2A - 1}{cot^2A + 1}\)

(i) tan2A = tan(A + B) =\(\frac{tanA + tanA}{1 - tanA. tanA}\) =\(\frac{2tanA}{1 - tan^2A}\)

(j) tan2A =\(\frac{2tanA}{1 - tan^2A}\) =\(\frac{\frac{2}{cotA}}{1 -\frac{1}{cot^2A}}\) =\(\frac{2}{cotA}\) x \(\frac{cot^2A}{cot^2A - 1}\)

(k) cot2A = cot (A + A) =\(\frac{cotA. cotA - 1}{cotA + cotA}\) =\(\frac{cot^2A - 1}{2cotA}\)

(l) cot2A =\(\frac{cot^2A - 1}{2cotA}\) =\(\frac{\frac{1}{tan^2A} - 1}{\frac{2}{tanA}}\) =\(\frac{1 - tan^2A}{tan^2A}\) x \(\frac{1 - tan^2 A}{2 tanA}\) =\(\frac{1 - tan^2 A}{2 tanA}\)

Some useful results

(a) 1 + cos2A = 1 + cos2A - sin2A = 1 - sin2A + cos2A = cos2A + cos2A = 2cos2A

(b) 1 - cos2A = 1 - (cos2A - sin2A) = 1 - cos2A + sin2A = sin2A + sin2A = 2sin2A

(c) 1 + sin2A = cos2A + sin2A + 2 sinA .cosA = cos2A + 2 cosA . sinA + sin2A = (cosA + sinA)2

(d) 1 - sin2A = cos2A + sin2A - 2sinA .cosA = cos2A - 2 cosA .sinA + sin2A = (cosA - sinA)2

Trigonometric ratios of 3A in terms of A

(a) sin3A

= sin(2A + A)

= sin2A. cosA + cos2A. sinA

= 2 sinA . cosA. cosA + (1 - 2sin2A). sinA

= 2 sinA (1 - sin2A) + sinA - 2sin\(^3\)A

= 2sinA. 2sin\(^3\)A + sinA - 2 sin\(^3\)A

= 3 sinA - 4 sin\(^3\)A

(b) cos3A

= cos (2a + A)

= cos2A. cosA - sin2A. sinA

= (2cos2A - 1) cosA - 2sinA. cosA. sinA

= 2cos\(^3\)A - cosA - 2cosA (1 - cos2A)

= 2 cos\(^3\)A - cosA - 2cosA + 2cos\(^3\)A

= 4 cos\(^3\)A - 3 cosA.

(c) tan3A

= tan( 2A + A)

=\(\frac{tan2A + tanA}{1 - tan2A. tanA}\)

=\(\frac{\frac{2 tan A}{1 - tan^2 A} + tanA}{1 -\frac{2tan A}{1 - tan^2 A}. tanA}\)

=\(\frac{2 tan A + tan A - tan^3 A}{1 - tan^2 A - 2 tan^2 A}\)

=\(\frac{3 tan A - tan^3 A}{1 - 3 tan^2 A}\)

Geometrical proof of 2A formulae :

.

Let, O be the centre of the circle ABC and AB be a diameter. Let ∠CAB = A.

Then, ∠COB = 2A

Here, ∠ACB = 900. Let CM is perpendicular to AB.

Then, ∠ACM = 900 - A and hence. ∠BCM = A.

Now,

sin2A =\(\frac{CM}{OC}\)

=\(\frac{2CM}{2OC}\)

=\(\frac{2CM}{AB}\)

= 2 \(\frac{CM}{AC}\). \(\frac{AC}{AB}\)

= 2sinA. cosA

And,

cos2A =\(\frac{OM}{OC}\)

=\(\frac{2OM}{2OC}\)

=\(\frac{2OM}{AB}\) 

= \(\frac{(AO + OM) - (AO - OM)}{AB}\)

=\(\frac{AM - BM}{AB}\) 

= \(\frac{AM}{AB}\) - \(\frac{BM}{AB}\)

= \(\frac{AM}{AC}\). \(\frac{AC}{AB}\) - \(\frac{BM}{BC}\). \(\frac{BC}{AB}\) 

= cosA. cosA - sinA. sinA

= cos2A - sin2A

Again,

tan2A = \(\frac{CM}{OM}\)

= \(\frac{2CM}{2OM}\)

=\(\frac{2CM}{(AO + OM) - (AO -OM)}\)

= \(\frac{2CM}{(AO + OM) - (BO - OM)}\)

=\(\frac{2CM}{AM - BM}\) 

=\(\frac{\frac{2CM}{AM}}{\frac{AM}{AM} - \frac{BM}{AM}}\)

=\(\frac{\frac{2CM}{AM}}{1 -\frac{BM}{CM}× \frac{CM}{AM}}\) 

=\(\frac{2 tan A}{1 - tan A. tan A}\) 

=\(\frac{2 tan A}{1 - tan^2 A}\)

Multiple angles formulae
sin2A = 2sinA. cosA tan2A =\(\frac{2cot A}{cot^2 A - 1}\)
sin2A =\(\frac{2tan A}{1 + tan^2 A}\) cot2A =\(\frac{cot^2 A - 1}{2cot A}\)
sin2A =\(\frac{2cot A}{1 + cot^2 A}\) cot2A =\(\frac{1 - tan^2 A}{2tan A}\)
cos2A = cos2A - sin2A 1 + cos2A = 2cos2A
cos2A = 1 - 2sin2A 1 - cos2A = 2 sin2A
cos2A = 2cos2A - 1 1 + sin2A = (cosA + sinA)2
cos2A =\(\frac{1 - tan^2 A}{1 + ttan^2 A}\) 1 - sin2A = (cosA - sinA)2
cos2A =\(\frac{cot^2 A - 1}{cot^2 A + 1}\) sin3A = 3 sinA - 4 sin \(^3\) A
tan2A =\(\frac{2 tan A}{1 - tan^2 A}\) cos3A = 4 cos\(^3\)A - 3cos A
tan3 A =\(\frac{3 tan A - tan^3 A}{1 - 3tan^2 A}\)  

 

Lesson

Trigonometry

Subject

Optional Mathematics

Grade

Grade 10

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