Equation or identity (1)
Which of these equations concerning the angles of triangles are always true?
Age
16 to 18
Challenge level
Problem
Throughout, $A$, $B$ and $C$ are the angles of a triangle.
For each of the following, decide whether it is an identity (true for all triangles) or an equation (there is a triangle for which it is not true).
If it is an identity, true for all triangles, then you should prove it (perhaps using trigonometric identities that you already know).
If it is an equation, then give an example of a triangle for which it is not true. You could also try to solve the equation (that is, find all triangles for which it is true).
1. $\sin(\frac{\pi}{2} - A) = \cos A$.
2. $\sec^2 A - \tan^2 A = 1$.
3. $\tan(A + B) = - \tan C$.
4. $\sin^2 A - \cos^2 B = 1$.
5. $\sin(\pi - A) = \sin B$.
6. $\sin^2 A - 3\cos A = $ cosec$^2 C - \cot^2 C$.
This is an Underground Mathematics resource.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Student Solutions
Here are Julian from British School Manila's solutions to parts 1,2,3 of the question
1. By the addition formulae we have that $\sin(\pi/2-A)=\sin(\pi/2)\cos(A)-\sin(A)\cos(\pi/2)=\cos(A)$ for all angles A, and hence this is an identity.
2. By pythagoras theorem we have that $o^2+a^2=h^2$, where o, a, and h are the opposite, adjacent and hypotenuse sides respectively of a right angled triangle with angle A. Dividing through by $a^2$ gives $\tan^2(A)+1=\sec^2(A)$. Rearranging gives the statement and hence this is also an identity.
3. Since A,B,and C are the angles of a triangle we know they are related by $A+B+C=\pi$, so $\tan(A+B)=\tan(\pi-C)=\tan(-C)=-\tan(C)$ for all possible values of A,B and C, and hence this is also an identity.
Here are Kyle from Kimberley STEM College's solutions to parts 4,5,6 of the question
4. Here $\cos^2(B)$ can not be negative and $\sin^2(A)$ cannot be greater than 1, so it can be concluded that $\cos^2(B)=0$ and $\sin^2(A)=1$. This does not hold for all angles A and B so we can already conclude at this point that the statement is an equation. In fact this equation can not be satisfied by any triangle because the angles of a triangle must satisfy $0 < A, B < \pi$ and so this would require $A=\pi/2$ and $B=\pi/2$ which is impossible in a triangle as it would mean $C=0$.
5. By the addition formulae have that $\sin(\pi-A)=\sin(\pi)\cos(A)-\sin(A)\cos(\pi)=\sin(A)$, and so the statement reduces to $\sin(A)=\sin(B)$. We can see at this point that this does not hold for all angles A and B, and so this is an equation. In the region $0 < A, B < \pi $ this is only satisfied when either $A=B$ or $B=\pi-A$. We can reject the latter solution because this cannot happen if A and B are angles of a triangle, so the solution is $A=B$
(So in summary this is only satisfied when A and B are the equal angles in an isosceles triangle).
6. We can simplify the right hand side as follows:
$RHS=\operatorname{cosec}^2(C)-\cot^2(C)=1/\sin^2(C)-1/\tan^2(C)=(1-\cos^2(C))/\sin^2(C)$
$=\sin^2(C)/\sin^2(C)=1$. Also we can write the left hand side entirely in terms of $\cos(A)$:
$LHS=\sin^2(A)-3\cos(A)=1-\cos^2(A)-3\cos(A)$. Hence the equation reduces to $\cos^2(A)+3\cos(A)=0$ which factorises to $\cos(A)(\cos(A)+3)=0$ and so we see that the only solutions are when $\cos(A)=0$ which, since A is an angle of a triangle, means that $A=\pi/2$. Since this doesn't hold for all A, it is an equation.