Category Archives: High School Trigonometry

The Proof of the Tangent Half-Angle Formula

Posted on May 3, 2012 by | 1 Comment

In this post, we prove the following trigonometric identity:

\displaystyle \tan \frac{\theta}{2} = \frac{\sin\theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}.

Proof

Consider a semi-circle with “center” O and diameter AB and radius equal to 1 unit as shown below.  If we let \angle BOC =\theta, then by the Inscribed Angle Theorem, \angle CAB = \frac{\theta}{2}.

Draw CD perpendicular to OB as shown in the second figure. We can compute for the sine and cosine of \theta which equal to the lengths of CD and OD, respectively. In effect, BD = 1 - \cos \theta and AD = 1 + \cos \theta. Continue reading →

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Posted in High School Mathematics, High School Trigonometry, Problem Solving and Proofs

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The Complements Theorem

Posted on September 2, 2011 by | Leave a comment

In this post, we discuss the proof behind one of the most commonly used identities in trigonometry. We examine the equations below and  show why the relationships always hold.

\sin \theta = \cos (90^{\circ} - \theta)
\cos \theta = \sin(90^{\circ} - \theta)

To students who have taken trigonometry, I’m sure that you have met these equation before.  The proof of these equations are as follows.

Consider triangle ABC right angled at C. From the definitions, we know that

\sin A = \displaystyle\frac{a}{c}
\cos A = \displaystyle\frac{b}{c}
\sin B = \displaystyle\frac{a}{c}
\cos B = \displaystyle\frac{a}{c}

 Therefore, (1) \sin A = \cos B and (2) \cos A = \sin B.

Now, If we let A = \theta, then B = 90^{\circ} - \theta, then substituting the values of A and B in (1) and (2), we have

\sin \theta = \cos (90^{\circ} - \theta)
\cos \theta = \sin (90^{\circ} - \theta)

 and these are what we want to show.

As exercises, use the strategy above, or any strategy you want to prove the following identities.

1.) \cot\theta = \tan (90^{\circ} - \theta)
2.) \sec \theta = \csc(90^{\circ} - \theta)

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An elementary proof of the cosine law

Posted on April 12, 2011 by | 4 Comments

The cosine law states that in triangle ABC with side lengths a, b, and c, the following equations are satisfied:

a ^2 = b^2 + c^2 - 2bc \cos A*

b ^2 = a^2 + c^2 - 2ac \cos B

c ^2 = a^2 + b^2 - 2ab \cos C

The discussion below shows how these equations were derived. Continue reading →

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Posted in High School Geometry, High School Mathematics, High School Trigonometry, Problem Solving and Proofs

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