# Is the substitution of standard angles while proving the equality of trigonometric formulas allowed?

Here is a problem that my class 10 maths teacher gave me:

Prove that $\sec^4\theta$ - $\sec^2\theta$ = $\tan^4\theta$ + $\tan^2\theta$

She expected me to use trigonometric identities to prove such equality, but I instead substituted $\theta$ with standard angles, and substituted that with their values.

Here's what I did:

$\sec^4\theta -\sec^2\theta = \tan^4\theta + \tan^2\theta$

take $\theta$ as $45^o$

$\sec^4 45^o - \sec^2 45^o =\tan^4 45^o + \tan^2 45^o$

$(\sqrt2)^4 - (\sqrt2)^2 = (1)^4 +(1)^2$

$4 - 2 = 1+1$

$2 = 2$

therefore LHS = RHS

Reason: Standard angles are universal truths, henceforth they work throughout the universe. Also, the functions used require a parameter (although not required in proving this using identities), so I substituted them all with $45^o$.

So, my question is 'Is it okay if I substitute standard angle values for trigonometric functions while proving their equality?'.

• You prove that this equality is valid for $\theta=45^\circ$; but what about other angles? – Michael Galuza Jul 22 '15 at 15:04
• It's called checking, not proof :) – Michael Galuza Jul 22 '15 at 15:08
• This has nothing to do with trigonometric functions per se. You cannot prove the identity $(x+y)^2=x^2+x+y+y^2$ by plugging in $x=y=1$ and checking whether you get a true statement. – Christian Blatter Jul 22 '15 at 15:18
• 1. What are "standard angles"? 2. What is "Standard angles are universal truths" supposed to mean? 3. "henceforth they work throughout the universe"??? – Daniel Fischer Jul 22 '15 at 15:24
• So are you saying that $\sin\theta=\cos\theta$? Since it is true for $\theta=45^\circ$ then it must be true for all angles, right? – Joel Reyes Noche Jul 23 '15 at 1:25

For the purpose of formal proofs, your method is not acceptable and rigorous enough.

For this proof, you need to start from either RHS or LHS and arrive at other side of the equality sign. You should prove it for any general angle say $\theta$ as this is more rigorous. If you show it for $45^\circ$, how would you know it applies for other angles?

Using your example, I'll start from LHS and arrive at RHS:

$$\text{LHS} =\sec^4\theta -\sec^2\theta$$ Note: $\sec^2\theta = \tan^2\theta + 1$

$$\text{LHS} = (\sec^2\theta)^2 -\sec^2\theta= (\tan^2\theta + 1)^2 -\tan^2\theta -1$$

$$\text{LHS}= \tan^4\theta+2\tan^2\theta -\tan^2\theta+1-1=\tan^4\theta +\tan^2\theta = \text{RHS}$$

That is one way to go through this proof and I'm sure there are other methods too, I hope it made sense and yes you have to use identities.

• thank you. please reread my post.i have added the reason – Switch Jul 22 '15 at 15:22
• @Switch, Great, pls see Daniel Fischer's comments on your post. – John_dydx Jul 22 '15 at 15:28

No. It is a good practise to examine some special cases, but that does not make a proof. To prove the statement using your substitution method, you would have to examine every possible value of $\theta$ which is impossible. Thus you have to use trignometric identities since they hold for general angle $\theta$.

• but standard angles should work for any of these, right? so as the values of those angles are universal truths they should be right, isnt it? – Switch Jul 22 '15 at 15:11
• Sorry, I don't understand your second sentence. What do you mean by "the values of those angles are universal truths"? – 3x89g2 Jul 22 '15 at 15:13
• the results of those angles are already proven. hence they work no matter what. also i missed out an important point to mention in the question. ill let u know after editing it. – Switch Jul 22 '15 at 15:16
• edited the post – Switch Jul 22 '15 at 15:21

HINT:

$$\sec^4A-\tan^4A=(\sec^2A)^2-(\tan^2A)^2=(\sec^2A-\tan^2A)(\sec^2A+\tan^2A)$$

$$\implies\sec^4A-\tan^4A=\sec^2A+\tan^2A$$

Your approach would work if you were able to prove rigorously that it is possible to extend the result from a single value to a domain. For instance suppose that you prove that a function never changes sign. If you compute the sign in a special (easy?) case and found out that it assumes a positive value, you have proved that the function is everywhere positive.

As it stands, you have proved that the equation holds true for a specific value, but you still miss the part where you prove that from this fact you can extend to the whole domain.

You use a similar type of reasoning with the proof by induction.

LHS: $$\sec^4\theta - \sec^2\theta = \sec^2\theta(\sec^2\theta - 1) = \frac{1}{\cos^2\theta}\frac{1-\cos^2\theta}{\cos^2\theta}=\frac{\sin^2\theta}{\cos^4\theta}$$ RHS: $$\tan^4\theta + \tan^2\theta = \tan^2\theta (1 + \tan^2\theta) = \frac{\sin^2\theta}{\cos^2\theta}\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta}=\frac{\sin^2\theta}{\cos^4\theta}$$ It's all.

• yes, she asked me to prove – Switch Jul 22 '15 at 15:23
• Your second equation is wrong. $\sec^2 \theta = 1/\cos^2 \theta$. – Zain Patel Jul 22 '15 at 15:25
• @ZainPatel, oops. I'll edit. My bad – Michael Galuza Jul 22 '15 at 15:29