# IBDP Maths analysis and approaches Topic: SL 3.6 :Pythagorean identities HL Paper 1

## Question

In the triangle ABC, $${\rm{A\hat BC}} = 90^\circ$$ , $${\text{AC}} = \sqrt {\text{2}}$$ and AB = BC + 1.

a.Show that cos $$\hat A – \sin \hat A = \frac{1}{{\sqrt 2 }}$$. [3]

b.By squaring both sides of the equation in part (a), solve the equation to find the angles in the triangle. [8]

c.Apply Pythagoras’ theorem in the triangle ABC to find BC, and hence show that $$\sin \hat A = \frac{{\sqrt 6 – \sqrt 2 }}{4}$$. [6]

d. Hence, or otherwise, calculate the length of the perpendicular from B to [AC]. [4]

## Markscheme

a.$$\cos \hat A = \frac{{{\text{BA}}}}{{\sqrt 2 }}$$     A1

$$\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }}$$     A1

$$\cos \hat A – \sin \hat A = \frac{{{\text{BA}} – {\text{BC}}}}{{\sqrt 2 }}$$     R1

$$= \frac{1}{{\sqrt 2 }}$$     AG

[3 marks]

b.

$${\cos ^2}\hat A – 2\cos \hat A\sin \hat A + {\sin ^2}\hat A = \frac{1}{2}$$     M1A1

$$1 – 2\sin \hat A\cos \hat A = \frac{1}{2}$$     M1A1

$$\sin 2\hat A = \frac{1}{2}$$     M1

$$2\hat A = 30^\circ$$     A1

angles in the triangle are 15° and 75°     A1A1

[8 marks]

c.

$${\text{B}}{{\text{C}}^2} + {({\text{BC}} + 1)^2} = 2$$     M1A1

$$2{\text{B}}{{\text{C}}^2} + 2{\text{BC}} – 1 = 0$$     A1

$${\text{BC}} = \frac{{ -2 + \sqrt {12} }}{4}\left( { = \frac{{\sqrt 3 – 1}}{2}} \right)$$     M1A1

$$\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }} = \frac{{\sqrt 3 – 1}}{{2\sqrt 2 }}$$     A1

$$= \frac{{\sqrt 6 – \sqrt 2 }}{4}$$     AG

[6 marks]

d.

EITHER

$$h = {\text{ABsin}}\hat A$$     M1

$$= ({\text{BC}} + 1)\sin \hat A$$     A1

$$= \frac{{\sqrt 3 + 1}}{2} \times \frac{{\sqrt 6 – \sqrt 2 }}{4} = \frac{{\sqrt 2 }}{4}$$     M1A1

OR

$$\tfrac{1}{2}AB.BC = \tfrac{1}{2}AC.h$$     M1

$$\frac{{\sqrt 3 – 1}}{2} \cdot \frac{{\sqrt {3 + 1} }}{2} = \sqrt {2h}$$     A1

$$\frac{2}{4} = \sqrt 2 h$$     M1

$$h = \frac{1}{{2\sqrt 2 }}$$     A1

[4 marks]

## Question

In the triangle ABC, $${\rm{A\hat BC}} = 90^\circ$$ , $${\text{AC}} = \sqrt {\text{2}}$$ and AB = BC + 1.

a.Show that cos $$\hat A – \sin \hat A = \frac{1}{{\sqrt 2 }}$$. [3]

b.By squaring both sides of the equation in part (a), solve the equation to find the angles in the triangle. [8]

c.Apply Pythagoras’ theorem in the triangle ABC to find BC, and hence show that $$\sin \hat A = \frac{{\sqrt 6 – \sqrt 2 }}{4}$$. [6]

d.Hence, or otherwise, calculate the length of the perpendicular from B to [AC]. [4]

## Markscheme

a. $$\cos \hat A = \frac{{{\text{BA}}}}{{\sqrt 2 }}$$     A1

$$\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }}$$     A1

$$\cos \hat A – \sin \hat A = \frac{{{\text{BA}} – {\text{BC}}}}{{\sqrt 2 }}$$     R1

$$= \frac{1}{{\sqrt 2 }}$$     AG

[3 marks]

b.

$${\cos ^2}\hat A – 2\cos \hat A\sin \hat A + {\sin ^2}\hat A = \frac{1}{2}$$     M1A1

$$1 – 2\sin \hat A\cos \hat A = \frac{1}{2}$$     M1A1

$$\sin 2\hat A = \frac{1}{2}$$     M1

$$2\hat A = 30^\circ$$     A1

angles in the triangle are 15° and 75°     A1A1

[8 marks]

c.

$${\text{B}}{{\text{C}}^2} + {({\text{BC}} + 1)^2} = 2$$     M1A1

$$2{\text{B}}{{\text{C}}^2} + 2{\text{BC}} – 1 = 0$$     A1

$${\text{BC}} = \frac{{ -2 + \sqrt {12} }}{4}\left( { = \frac{{\sqrt 3 – 1}}{2}} \right)$$     M1A1

$$\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }} = \frac{{\sqrt 3 – 1}}{{2\sqrt 2 }}$$     A1

$$= \frac{{\sqrt 6 – \sqrt 2 }}{4}$$     AG

[6 marks]

d.

EITHER

$$h = {\text{ABsin}}\hat A$$     M1

$$= ({\text{BC}} + 1)\sin \hat A$$     A1

$$= \frac{{\sqrt 3 + 1}}{2} \times \frac{{\sqrt 6 – \sqrt 2 }}{4} = \frac{{\sqrt 2 }}{4}$$     M1A1

OR

$$\tfrac{1}{2}AB.BC = \tfrac{1}{2}AC.h$$     M1

$$\frac{{\sqrt 3 – 1}}{2} \cdot \frac{{\sqrt {3 + 1} }}{2} = \sqrt {2h}$$     A1

$$\frac{2}{4} = \sqrt 2 h$$     M1

$$h = \frac{1}{{2\sqrt 2 }}$$     A1

[4 marks]

## Question

Show that $$\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \sec 2A + \tan 2A$$ .

## Markscheme

METHOD 1

$$\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \sec 2A + \tan 2A$$

consider right hand side

$$\sec 2A + \tan 2A = \frac{1}{{\cos 2A}} + \frac{{\sin 2A}}{{\cos 2A}}$$     M1A1

$$= \frac{{{{\cos }^2}A + 2\sin A\cos A + {{\sin }^2}A}}{{{{\cos }^2}A – {{\sin }^2}A}}$$     A1A1

Note: Award A1 for recognizing the need for single angles and A1 for recognizing $${\cos ^2}A + {\sin ^2}A = 1$$ .

$$= \frac{{{{(\cos A + \sin A)}^2}}}{{(\cos A + \sin A)(\cos A – \sin A)}}$$     M1A1

$$= \frac{{\cos A + \sin A}}{{\cos A – \sin A}}$$     AG

METHOD 2

$$\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \frac{{{{(\cos A + \sin A)}^2}}}{{(\cos A + \sin A)(\cos A – \sin A)}}$$     M1A1

$$= \frac{{{{\cos }^2}A + 2\sin A\cos A + {{\sin }^2}A}}{{{{\cos }^2}A – {{\sin }^2}A}}$$     A1A1

Note: Award A1 for correct numerator and A1 for correct denominator.

$$= \frac{{1 + \sin 2A}}{{\cos 2A}}$$     M1A1

$$= \sec 2A + \tan 2A$$     AG

[6 marks]

## Question

a. Show that $$\cot \alpha = \tan \left( {\frac{\pi }{2} – \alpha } \right)$$ for $$0 < \alpha < \frac{\pi }{2}$$. [1]

b. Hence find $$\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x,{\text{ }}0 < \alpha < \frac{\pi }{2}}$$. [4]

## Markscheme

EITHER

a.use of a diagram and trig ratios

eg,

$$\tan \alpha = \frac{O}{A} \Rightarrow \cot \alpha = \frac{A}{O}$$

from diagram, $$\tan \left( {\frac{\pi }{2} – \alpha } \right) = \frac{A}{O}$$     R1

OR

use of $$\tan \left( {\frac{\pi }{2} – \alpha } \right) = \frac{{\sin \left( {\frac{\pi }{2} – \alpha } \right)}}{{\cos \left( {\frac{\pi }{2} – \alpha } \right)}} = \frac{{\cos \alpha }}{{\sin \alpha }}$$     R1

THEN

$$\cot \alpha = \tan \left( {\frac{\pi }{2} – \alpha } \right)$$     AG

[1 mark]

a.

$$\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x} = [\arctan x]_{\tan \alpha }^{\cot \alpha }$$     (A1)

Note:     Limits (or absence of such) may be ignored at this stage.

$$= \arctan (\cot \alpha ) – \arctan (\tan \alpha )$$     (M1)

$$= \frac{\pi }{2} – \alpha – \alpha$$     (A1)

$$= \frac{\pi }{2} – 2\alpha$$     A1

[4 marks]

### Question

Given that $$sin(x+\frac{\pi}{3})=2cos(x-\frac{\pi}{6})$$, find the values of tan x, cot x and cos x.

$$sin(x+\frac{\pi}{3})=2cons(x-\frac{\pi}{6} \Rightarrow sin x cos\frac{\po}{3}+sin\frac{\pi}{3}cos x=2(cosxcox\frac{\pi}{6}+sinxsin\frac{\pi}{6}$$
$$\Rightarrow \frac{1}{2}sinxcos\frac{\pi}{3}+\frac{\sqrt{3}}{2}cosx=2(\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx)$$
$$\Rightarrow sinx +\sqrt{3}cos x =2\sqrt{3}cos x+2sinx$$
$$\Rightarrow -sinx =\sqrt{3}cosx\Rightarrow tanx =-\sqrt{3}$$
$$ctx=-\frac{1}{\sqrt{3}}$$
$$tan^2x+1=\frac{1}{cos^2x}\Rightarrow cos^2x=\frac{1}{4}\Rightarrow tan x =-\sqrt{3}$$
(indeed, the original equations has solution $$-\frac{\pi}{3}+2k\pi,\frac{2\pi}{3}+2k\pi$$ and hence $$cos x = \pm \frac{1}{2})$$