Video Transcript
Simplify two sin squared 𝜃 plus
cos squared 𝜃 plus one over sec squared 𝜃.
To simplify this expression, which
is the sum of various trigonometric functions, means that we should be able to
combine some of these terms together or perhaps write the whole sum in terms of one
single trigonometric function. Let’s begin by looking at the final
term in this sum, which is one over sec squared of 𝜃.
Now, we recall first of all that
sec of 𝜃 is one of our reciprocal trigonometric functions. The definition of sec of 𝜃 is one
over cos of 𝜃. The one over is what makes it a
reciprocal. So sec squared of 𝜃, which just
means sec of 𝜃 all squared, is equal to one over cos 𝜃 all squared. And squaring the numerator and
denominator of this fraction separately, we see that sec squared of 𝜃 is equal to
one over cos squared of 𝜃.
Now, this final term isn’t sec
squared of 𝜃. It’s one over sec squared of
𝜃. So this means one divided by what
we’ve just shown sec squared of 𝜃 should be equal to. That’s one divided by one over cos
squared of 𝜃.
We recall that, to divide by a
fraction, we flip or invert that fraction. So its numerator becomes the
denominator. And the denominator becomes the
numerator. And then, we multiply instead of
divide. So one divided by one over cos
squared 𝜃 is equal to one multiplied by cos squared 𝜃 over one. This order simplifies to cos
squared 𝜃. So the final sum in our expression
is just cos squared 𝜃.
We can therefore replace this third
term in the sum with cos squared 𝜃. And we have that this sum is equal
to two sin squared 𝜃 plus cos squared 𝜃 plus another lot of cos squared 𝜃. These two lots of cos squared 𝜃
can be combined together to give two cos squared 𝜃. So the sum simplifies to two sin
squared 𝜃 plus two cos squared 𝜃.
Now we observe that the two terms
have a common factor of two. So we can factorise this sum by
two. This gives two multiplied by sin
squared 𝜃 plus cos squared 𝜃. Now at this point, we should
recognise that we can use one of our trigonometric identities, which is that, for
any angle 𝜃, sin squared of 𝜃 plus cos squared of 𝜃 is always equal to one.
We can see this using the unit
circle. Sin of 𝜃 and cos of 𝜃 give the
vertical and horizontal length in a right-angle triangle with a hypotenuse of
one. And by applying the Pythagorean
theorem, we see that sin squared of 𝜃 plus cos squared of 𝜃 is always equal to
one. So by replacing the sum in the
bracket with its value of one, we have two multiplied by one, which is equal to
two.
We’ve shown then that the sum, two
sin squared 𝜃 plus cos squared 𝜃 plus one over sec squared 𝜃, in fact just
simplifies to the number two.
